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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

Edited By Ramraj Saini | Updated on Sep 25, 2023 07:42 PM IST

Limits and Derivatives Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives are provided here. This Class 11 NCERT syllabus introduces an important area of mathematics called calculus. This is a branch of mathematics which deal with the study of change in the value of a function as the points in the domain change. In this article, you will get limits and derivatives class 11 NCERT solutions that are prepared by highly qualified teachers keeping in mind latest syllabus of CBSE 2023, simple and provide step by step solution to each problem. Students can get all NCERT solutions at one place and practice them to develop indepth understanding of the concepts.

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This Story also Contains
  1. Limits and Derivatives Class 11 Questions And Answers
  2. Limits and Derivatives Class 11 Questions And Answers PDF Free Download
  3. Limits and Derivatives Class 11 Solutions - Important Formulae
  4. Limits and Derivatives Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. limits and derivatives ncert solutions - Topics
  6. NCERT Solutions for Class 11 Mathematics - Chapter Wise
  7. Key Features Of Limits And Derivatives Class 11 Solutions
  8. NCERT Solutions for Class 11- Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives
NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

In class 11 maths chapter 13 question answer, you will get questions related to all the above topics. This chapter is very important for both class 11 final examination and various competitive exams like JEE Main, JEE Advanced, VITEEE, BITSAT, etc. There are 43 questions in 2 exercises limits and derivatives class 11. The first exercise of this chapter deals with problems on limits and the second exercise deals with problems on derivatives. Students can find all NCERT solution for class 11 at one pace for hare and practice them.

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Limits and Derivatives Class 11 Questions And Answers PDF Free Download

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Limits and Derivatives Class 11 Solutions - Important Formulae

Limits:

The limit of a function f(x) as x approaches a, denoted as lim(x→a) f(x), exists when both the left-hand limit (lim(x→a-) f(x)) and right-hand limit (lim(x→a+) f(x)) exist and are equal (lim(x→a-) f(x) = lim(x→a+) f(x)).

Left-Hand and Right-Hand Limits:

Left-Hand Limit: f(a-0) = lim(x→a-) f(x) = lim(h→0) f(a-h)

Right-Hand Limit: f(a+0) = lim(x→a+) f(x) = lim(h→0) f(a+h)

Properties of Limits:

If lim(x→a) f(x) and lim(x→a) g(x) both exist:

  • Sum Rule: lim(x→a) [f(x) ± g(x)] = lim(x→a) f(x) ± lim(x→a) g(x)

  • Constant Rule: lim(x→a) kf(x) = k lim(x→a) f(x)

  • Product Rule: lim(x→a) [f(x) * g(x)] = lim(x→a) f(x) * lim(x→a) g(x)

  • Quotient Rule: lim(x→a) [f(x)/g(x)] = [lim(x→a) f(x)] / [lim(x→a) g(x)] (provided lim(x→a) g(x) ≠ 0)

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Standard Limits:

  • lim(x→a) [(xn - an)/(x - a)] = na(n-1)

  • lim(x→0) [(sin(x))/x] = 1

  • lim(x→0) [(tan(x))/x] = 1

  • lim(x→0) [(ax - 1)/x] = ln(a)

  • lim(x→0) [(ex - 1)/x] = 1

  • lim(x→0) [(ln(1+x))/x] = 1

Derivatives:

The derivative of a function f at a point x, denoted as f'(x), is defined as:

  • f'(x) = lim(h→0) [(f(x+h) - f(x))/h]

Properties of Derivatives:

If f(x) and g(x) have derivatives in a common domain:c

  • Sum Rule: [d/dx (f(x) + g(x))] = [d/dx (f(x))] + [d/dx (g(x))]

  • Difference Rule: [d/dx (f(x) - g(x))] = [d/dx (f(x))] - [d/dx (g(x))]

  • Product Rule: [d/dx (f(x) * g(x))] = [d/dx f(x)] * g(x) + f(x) * [d/dx g(x)]

  • Quotient Rule: [d/dx (f(x)/g(x))] = ([d/dx f(x)] * g(x) - f(x) * [d/dx g(x)]) / [g(x)]2

Standard Derivatives:

  • [d/dx (x^n)] = n * x(n-1)

  • [d/dx (sin(x))] = cos(x)

  • [d/dx (cos(x))] = -sin(x)

  • [d/dx (tan(x))] = sec2(x)

  • [d/dx (cot(x))] = -csc2(x)

  • [d/dx (sec(x))] = sec(x) * tan(x)

  • [d/dx (csc(x))] = -csc(x) * cot(x)

  • [d/dx (a^x)] = ax * ln(a)

  • [d/dx (e^x)] = ex

  • [d/dx (ln(x))] = 1/x

Free download NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives for CBSE Exam.

Limits and Derivatives Class 11 NCERT Solutions (Intext Questions and Exercise)

Limits and derivatives class 11 questions and answers - Exercise: 13.1

Question:1 Evaluate the following limits

limx3x+3

Answer:

limx3x+3

limx33+3

6 Answer

Question:2 Evaluate the following limits limxπ(x22/7)

Answer:

Below you can find the solution:

limxπ(x22/7)=π227

Question:3 Evaluate the following limits limr1πr2

Answer:

The limit

limr1πr2=π(1)2=π

Answer is π

Question:4 Evaluate the following limits limx44x+3x2

Answer:

The limit

limx44x+3x2

4(4)+3(4)2

192 (Answer)

Question:5 Evaluate the following limits limx1x10+x5+1x1

Answer:

The limit

limx4x10+x5+1x1

(1)10+(1)5+1(1)1

11+12

12 (Answer)

Question:6 Evaluate the following limits limx0(x+1)51x

Answer:

The limit

limx0(x+1)51x

Lets put

x+1=y

since we have changed the function, its limit will also change,

so

x0,y0+1=1

So our function have became

limy1y51y1

Now As we know the property

limx1x5anxa=nan1


limy1y51y1=5(1)5=5

Hence,

limx0(x+1)51x=5

Question:7 Evaluate the following limits limx23x2x10x24

Answer:

The limit

limx23x2x10x24

limx2(x2)(3x+5)(x2)(x+2)

limx2(3x+5)(x+2)

(3(2)+5)((2)+2)

114 (Answer)

Question:8 Evaluate the following limits limx3x4812x25x3

Answer:

The limit

limx3x4812x25x3

At x = 2 both numerator and denominator becomes zero, so lets factorise the function

limx3(x3)(x+3)(x2+9)(x3)(2x+1)

limx3(x+3)(x2+9)(2x+1)

Now we can put the limit directly, so

limx3(x+3)(x2+9)(2x+1)

((3)+3)((3)2+9)(2(3)+1)

6×187

1087

Question:9 Evaluate the following limits limx0ax+bcx+1

Answer:

The limit,

limx0ax+bcx+1

a(0)+bc(0)+1

b (Answer)

Question:10 Evaluate the following limits limz1z1/31z1/61

Answer:

The limit

limz1z1/31z1/61

Here on directly putting limit , both numerator and the deniminator becomes zero so we factorize the function and then put the limit.

limz1z1/31z1/61=limz1z(1/6)212z1/61

=limz1(z(1/6)1)(z(1/6)+1)z1/61

\\=\lim_{z\rightarrow 1}(z^{1/6}+1)\\=(1^{1/6}+1)

=1+1

=2 (Answer)

Question:11 Evaluate the following limits limx1ax2+bx+ccx2+bx+a,a+b+c0

Answer:

The limit:

limx1ax2+bx+ccx2+bx+a,a+b+c0

Since Denominator is not zero on directly putting the limit, we can directly put the limits, so,

limx1ax2+bx+ccx2+bx+a,a+b+c0

=a(1)2+b(1)+cc(1)2+b(1)+a,

=a+b+ca+b+c,

=1 (Answer)

Question:12 Evaluate the following limits limx21x+x2x+2

Answer:

limx21x+x2x+2

Here, since denominator becomes zero on putting the limit directly, so we first simplify the function and then put the limit,

limx21x+x2x+2

=limx2x+22xx+2

=limx212x

=12(2)

=14 (Answer)

Question:13 Evaluate the following limits limx0sinaxbx

Answer:

The limit

limx0sinaxbx

Here on directly putting the limits, the function becomes 00 form. so we try to make the function in the form of sinxx . so,

limx0sinaxbx

=limx0sinax(ax)bx(ax)

=limx0sinaxaxab

As limx0sinxx=1

=1ab

=ab (Answer)

Question:14 Evaluate the following limits limx0sinaxsinbx,a,b0

Answer:

The limit,

limx0sinaxsinbx,a,b0

On putting the limit directly, the function takes the zero by zero form So,we convert it in the form of sinaa .and then put the limit,

limx0sinaxaxsinbxbxaxbx

=limax0sinaxaxlimbx0sinbxbxab

=ab (Answer)

Question:15 Evaluate the following limits limxπsin(πx)π(πx)

Answer:

The limit

limxπsin(πx)π(πx)

limxπsin(πx)(πx)×1π

=1×1π

=1π (Answer)

Question:16 Evaluate the following limits limx0cosxπx

Answer:

The limit

limx0cosxπx

the function behaves well on directly putting the limit,so we put the limit directly. So.

limx0cosxπx

=cos(0)π(0)

=1π (Answer)

Question:17 Evaluate the following limits limx0 cos2x1cosx1

Answer:

The limit:

limx0 cos2x1cosx1

The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit

limx0 cos2x1cosx1

limx0 2(sin2x)2(sin2(x2))

=limx0 (sin2x)x2(sin2(x2))(x2)2×x2(x2)2

=1212×4

=4 (Answer)

Question:18 Evaluate the following limits limx0ax+xcosxbsinx

Answer:

limx0ax+xcosxbsinx

The function takes the form zero by zero when we put the limit directly in the function,. since function consist of sin function and cos function, we try to make the function in the form of sinxx as we know that it tends to 1 when x tends to 0.

So,

limx0ax+xcosxbsinx

=1blimx0x(a+cosx)sinx

=1blimx0xsinx×(a+cosx)

=1b×1×(a+cos(0))

=a+1b (Answer)

Question:19 Evaluate the following limits limx0xsecx

Answer:

limx0xsecx

As function doesn't create any abnormality on putting the limit directly,we can put limit directly. So,

limx0xsecx

=(0)×1

=0 . (Answer)

Question:20 Evaluate the following limits limx0sinax+bxax+sinbxa,b,a+b0

Answer:

limx0sinax+bxax+sinbxa,b,a+b0

The function takes the zero by zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So

limx0sinax+bxax+sinbxa,b,a+b0

=limx0sinaxaxax+bxax+sinbxbxbx

=limx0sinaxaxa+ba+sinbxbxb

=1a+ba+1b

=a+ba+b

=1 (Answer)

Question:21 Evaluate the following limits limx0(cscxcotx)

Answer:

limx0(cscxcotx)

On putting the limit directly the function takes infinity by infinity form, So we simplify the function and then put the limit

limx0(cscxcotx)

=limx0(1sinxcosxsinx)

=limx0(1cosxsinx)

=limx0(2sin2(x2)sinx)

=limx0(2sin2(x2)(x2)2)((x2)2sinx)

=limx024(sin2(x2)(x2)2)((x)sinx)x

=24×(1)2×0

=0 (Answer)

Question:22 Evaluate the following limits limxπ/2tan2xxπ/2

Answer:

limxπ/2tan2xxπ/2

The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits,

So

Let's put

y=xπ2

Since we are changing the variable, limit will also change.

as

xπ2,y=xπ2π2π2=0

So function in new variable becomes,

limy0tan2(y+π2)y+π/2π/2

=limy0tan(2y+π)y

As we know tha property tan(π+x)=tanx

=limy0tan(2y)y

=limy0sin2y2y2cos2y

=1×2

=2 (Answer)

Question:23 Find limx0f(x)limx1f(x)wheref(x)={2x+3x03(x+1)x>0

Answer:

Given Function

f(x)={2x+3x03(x+1)x>0

Now,

Limit at x = 0 :

atx=0

: limx0f(x)=limx0(2x+3)=2(0)+3=3

atx=0+

limx0+f(x)=limx0+3(x+1)=3(0+1)=3

Hence limit at x = 0 is 3.

Limit at x = 1

atx=1+

limx1+f(x)=limx1+3(x+1)=3(1+1)=6

atx=1

limx1f(x)=limx13(x+1)=3(1+1)=6

Hence limit at x = 1 is 6.

Question:24 Find limx1f(x),wheref(x)={x21x0x21x>1

Answer:

limx1f(x),wheref(x)={x21x0x21x>1

Limit at x=1+

limx1+f(x)=limx1(x21)=(1)21=2

Limit at x=1

limx1f(x)=limx1(x21)=(1)21=0

As we can see that Limit at x=1+ is not equal to Limit at x=1 ,The limit of this function at x = 1 does not exists.

Question:25 Evaluate limx0f(x),wheref(x)={|x|xx00x=0

Answer:

limx0f(x),wheref(x)={|x|xx00x=0

The right-hand Limit or Limit at x=0+

limx0+f(x)=limx0+|x|x=xx=1

The left-hand limit or Limit at x=0

limx0f(x)=limx0|x|x=xx=1

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Question:26 Evaluate limx0f(x),wheref(x)={x|x|x00x=0

Answer:

limx0f(x),wheref(x)={x|x|x00x=0

The right-hand Limit or Limit at x=0+

limx0+f(x)=limx0+x|x|=xx=1

The left-hand limit or Limit at x=0

limx0f(x)=limx0x|x|=xx=1

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Question:27 Find limx5f(x),wheref(x)=|x|5

Answer:

limx5f(x),wheref(x)=|x|5

The right-hand Limit or Limit at x=5+

limx5+f(x)=limx5+|x|5=55=0

The left-hand limit or Limit at x=5

limx5f(x)=limx5|x|5=55=0

Since Left-hand limit and right-hand limit are equal, The limit of this function at x = 5 is 0.

Question:28 Suppose

f(x)={a+bxx<14x=1baxx>1 f (x) = f (1) what are possible values of a and b?

Answer:

Given,

f(x)={a+bxx<14x=1baxx>1

And

limx1f(x)=f(1)

Since the limit exists,

left-hand limit = Right-hand limit = f(1).

Left-hand limit = f(1)

limx1f(x)=limx1(a+bx)=a+b(1)=a+b=4

Right-hand limit

limx1+f(x)=limx1(bax)=ba(1)=ba=4

From both equations, we get that,

a=0 and b=4

Hence the possible value of a and b are 0 and 4 respectively.

Question:29 Let a1, a2, . . ., an be fixed real numbers and define a function f(x)=(xa1)(xa2)...(xan). What is limxa1 f (x) ? For some aa1,a2....an , compute l limxaf(x)

Answer:

Given,

f(x)=(xa1)(xa2)...(xan).

Now,

limxa1f(x)=limxa1[(xa1)(xa2)...(xan)].=[limxa1(xa1)][limxa1(xa2)][limxa1(xan)]=0

Hence

, limxa1f(x)=0

Now,

limxaf(x)=limxa(xa1)(xa2)...(xan)

limxaf(x)=(aa1)(aa2)(aa3)

Hence

limxaf(x)=(aa1)(aa2)(aa3) .

Question:30 If f(x)={|x|+1x<00x=0|x|1x>0 For what value (s) of a does limxaf(x) exists ?

Answer:

f(x)={|x|+1x<00x=0|x|1x>0

Limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,

Case 1: when a = 0

The right-hand Limit or Limit at x=0+

limx0+f(x)=limx0+|x|1=11=0

The left-hand limit or Limit at x=0

limx0f(x)=limx0|x|+1=0+1=1

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Case 2: When a < 0

The right-hand Limit or Limit at x=a+

limxa+f(x)=limxa+|x|1=a1

The left-hand limit or Limit at x=a

limxaf(x)=limxa|x|1=a1

Since LHL = RHL, Limit exists at x = a and is equal to a-1.

Case 3: When a > 0

The right-hand Limit or Limit at x=a+

limxa+f(x)=limxa+|x|+1=a+1

The left-hand limit or Limit at x=a

limxaf(x)=limxa|x|+1=a+1

Since LHL = RHL, Limit exists at x = a and is equal to a+1

Hence,

The limit exists at all points except at x=0.

Question:31 If the function f(x) satisfies limx1f(x)2x21=π , evaluate limx1f(x)

Answer:

Given

limx1f(x)2x21=π

Now,

limx1f(x)2x21=limx1(f(x)2)limx1(x21)=π

limx1(f(x)2)=πlimx1(x21)

limx1(f(x)2)=π(11)

limx1(f(x)2)=0

limx1f(x)=2

Question:32 If

\\f(x)=\begin{bmatrix} mx^2+n\ \ ;x<0 & \\nx+m\ \ ;x \leq0\leq 1t &\\ nx^3+m\ \ ;x>1 \end{bmatrix}

Answer:

Given,

\\f(x)=\begin{bmatrix} mx^2+n\ \ ;x<0 & \\nx+m\ \ ;x \leq0\leq 1t &\\ nx^3+m\ \ ;x>1 \end{bmatrix}

Case 1: Limit at x = 0

The right-hand Limit or Limit at x=0+

limx0+f(x)=limx0+nx+m=n(0)+m=m

The left-hand limit or Limit at x=0

limx0f(x)=limx0mx2+n=m(0)2+n=n

Hence Limit will exist at x = 0 when m = n .

Case 2: Limit at x = 1

The right-hand Limit or Limit at x=1+

limx1+f(x)=limx1+nx3+m=n(1)3+m=n+m

The left-hand limit or Limit at x=1

limx1f(x)=limx1nx+m=n(1)+m=n+m

Hence Limit at 1 exists at all integers.

Class 11 maths chapter 13 NCERT solutions - Exercise: 13.2

Question:1 Find the derivative of x22atx=10

Answer:

F(x)= x22

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

The derivative of f(x) at x = 10:

f(10)=limh0f(10+h)f(10)h

f(10)=limh0(10+h)22((10)22)h

f(10)=limh0100+20h+h22100+2h

f(10)=limh020h+h2h

f(10)=limh020+h

f(10)=20+0

f(10)=20

Question:2 Find the derivative of x at x = 1.

Answer:

Given

f(x)= x

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

The derivative of f(x) at x = 1:

f(1)=limh0f(1+h)f(1)h

f(1)=limh0(1+h)(1)h

f(1)=limh0(h)h

f(1)=1 (Answer)

Question:3 Find the derivative of 99x at x = l00.

Answer:

f(x)= 99x

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

The derivative of f(x) at x = 100:

f(100)=limh0f(100+h)f(100)h

f(100)=limh099(100+h)99(100)h

f(100)=limh099hh

f(100)=99

Question:4 (i) Find the derivative of the following functions from first principle. x327

Answer:

Given

f(x)= x327

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h


f(x)=limh0(x+h)327((x)327)h

f(x)=limh0x3+h3+3x2h+3h2x27+x3+27h

f(x)=limh0h3+3x2h+3h2xh

f(x)=limh0h2+3x2+3hx

f(x)=3x2

Question:4.(ii) Find the derivative of the following function from first principle. (x1)(x2)

Answer:

f(x)= (x1)(x2)

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0(x+h1)(x+h2)(x1)(x2)h

f(x)=limh0x2+xh2x+hx+h22hxh+2x2+2x+x2h

f(x)=limh02hx+h23hh

f(x)=limh02x+h3

f(x)=2x3 (Answer)

Question:4(iii) Find the derivative of the following functions from first principle. 1/x2

Answer:

f(x)= 1/x2

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh01/(x+h)21/(x2)h

f(x)=limh0x2(x+h)2(x+h)2x2h

f(x)=limh0x2x22xhh2h(x+h)2x2

f(x)=limh02xhh2h(x+h)2x2

f(x)=limh02xh(x+h)2x2

f(x)=2x0(x+0)2x2

f(x)=2x3 (Answer)

Question:4(iv) Find the derivative of the following functions from first principle. x+1x1

Answer:

Given:

f(x)=x+1x1

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0x+h+1x+h1x+1x1h

f(x)=limh0(x+h+1)(x1)(x+1)(x+h1)(x1)(x+h1)h

f(x)=limh0x2x+hxh+x1x2xh+xxh+1(x1)(x+h1)h

f(x)=limh02h(x1)(x+h1)h

f(x)=limh02(x1)(x+h1)

f(x)=2(x1)(x+01)

f(x)=2(x1)2

Question:5 For the function f(x)=x100100+x9999+....+x22+x+1 Prove that f '(1) =100 f '(0).

Answer:

f(x)=x100100+x9999+....+x22+x+1

As we know, the property,

f(xn)=nxn1

applying that property we get

f(x)=100x99100+99x9899+....+2x2+1+0

f(x)=x99+x98+......x+1

Now.

f(0)=099+098+......0+1=1

f(1)=199+198+......1+1=100

So,

Here

1×100=100

f(0)×100=f(1)

Hence Proved.

Question:6 Find the derivative of xn+axn1+a2xn2+....+an1x+an for some fixed real number a.

Answer:

Given

f(x)=xn+axn1+a2xn2+....+an1x+an

As we know, the property,

f(xn)=nxn1

applying that property we get

f(x)=nxn1+a(n1)xn2+a2(n2)xn3+....+an11+0

f(x)=nxn1+a(n1)xn2+a2(n2)xn3+....+an1

Question:7(i) For some constants a and b, find the derivative of (xa)(xb)

Answer:

Given

f(x)=(xa)(xb)=x2axbx+ab

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=2xab

Question:7(ii) For some constants a and b, find the derivative of (ax2+b)2

Answer:

Given

f(x)=(ax2+b)2=a2x4+2abx2+b2

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying those properties we get

f(x)=4a2x3+2(2)abx+0

f(x)=4a2x3+4abx

f(x)=4ax(ax2+b)

Question:7(iii) For some constants a and b, find the derivative of xaxb

Answer:

Given,

f(x)=xaxb

Now As we know the quotient rule of derivative,

d(y1y2)dx=y2dy1dxy1dy2dxy22

So applying this rule, we get

d(xaxb)dx=(xb)d(xa)dx(xa)d(xb)dx(xb)2

d(xaxb)dx=(xb)(xa)(xb)2

d(xaxb)dx=ab(xb)2

Hence

f(x)=ab(xb)2

Question:8 Find the derivative of xnanxa for some constant a.

Answer:

Given,

f(x)=xnanxa

Now As we know the quotient rule of derivative,

d(y1y2)dx=y2dy1dxy1dy2dxy22

So applying this rule, we get

d(xnanxa)dx=(xa)d(xnan)dx(xnan)d(xa)dx(xa)2

d(xnanxa)dx=(xa)nxn1(xnan)(xa)2

d(xnanxa)dx=nxnanxn1xn+an(xa)2

Hence

f(x)=nxnanxn1xn+an(xa)2

Question:9(i) Find the derivative of 2x3/4

Answer:

Given:

f(x)=2x3/4

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=20

f(x)=2

Question:9(ii) Find the derivative of (5x3+3x1)(x1)

Answer:

Given.

f(x)=(5x3+3x1)(x1)=5x4+3x2x5x33x+1

f(x)=5x45x3+3x24x+1

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=5(4)x35(3)x2+3(2)x4+0

f(x)=20x315x2+6x4

Question:9(iii) Find the derivative of x3(5+3x)

Answer:

Given

f(x)=x3(5+3x)=5x3+3x2

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=(3)5x4+3(2)x3

f(x)=15x46x3

Question:9(iv) Find the derivative of x5(36x9)

Answer:

Given

f(x)=x5(36x9)=3x56x4

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=(5)3x46(4)x5

f(x)=15x4+24x5

Question:9(v) Find the derivative of x4(34x5)

Answer:

Given

f(x)=x4(34x5)=3x44x9

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=(4)3x5(9)4x10

f(x)=12x5+36x10

Question:9(vi) Find the derivative of 2x+1x23x1

Answer:

Given

f(x)=2x+1x23x1

As we know the quotient rule of derivative:

d(y1y2)dx=y2dy1dxy1dy2dxy22

and the property

d(y1+y2)dx=dy1dx+dy2dx

So applying this rule, we get

d(2x+1x23x1)dx=(x+1)d(2)dx2d(x+1)dx(x+1)2(3x1)d(x2)dxx2d(3x1)dx(3x1)2

d(2x+1x23x1)dx=2(x+1)2(3x1)2xx23(3x1)2

d(2x+1x23x1)dx=2(x+1)26x22x3x2(3x1)2

d(2x+1x23x1)dx=2(x+1)23x22x(3x1)2

Hence

f(x)=2(x+1)23x22x(3x1)2

Question:10 Find the derivative of cosx from first principle.

Answer:

Given,

f(x)= cosx

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0cos(x+h)cos(x)h

f(x)=limh0cos(x)cos(h)sin(x)sin(h)cos(x)h

f(x)=limh0cos(x)cos(h)cos(x)hsin(x)sin(h)h

f(x)=limh0cos(x)(cos(h)1)hsin(x)sin(h)h

f(x)=limh0cos(x)2sin2(h/2)hsin(x)sinhh

f(x)=cos(x)(0)sinx(1)

f(x)=sin(x)

Question:11(i) Find the derivative of the following functions: sinxcosx

Answer:

Given,

f(x)= sinxcosx

Now, As we know the product rule of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

So, applying the rule here,

d(sinxcosx)dx=sinxdcosxdx+cosxdsinxdx

d(sinxcosx)dx=sinx(sinx)+cosx(cosx)

d(sinxcosx)dx=sin2x+cos2x

d(sinxcosx)dx=cos2x

Question:11(ii) Find the derivative of the following functions: secx

Answer:

Given

f(x)=secx=1cosx

Now As we know the quotient rule of derivative,

d(y1y2)dx=y2dy1dxy1dy2dxy22

So applying this rule, we get

d(1cosx)dx=cosxd(1)dx1d(cosx)dxcos2x

d(1cosx)dx=1(sinx)cos2x

d(1cosx)dx=sinxcos2x=sinxcosx1cosx

d(secx)dx=tanxsecx

Question:11 (iii) Find the derivative of the following functions: 5secx+4cosx

Answer:

Given

f(x)=5secx+4cosx

As we know the property

d(y1+y2)dx=dy1dx+dy2dx

Applying the property, we get

d(5secx+4cosx)dx=d(5secx)dx+d(4cosx)dx

d(5secx+4cosx)dx=5tanxsecx+4(sinx)

d(5secx+4cosx)dx=5tanxsecx4sinx

Question:11(iv) Find the derivative of the following functions: cscx

Answer:

Given :

f(x)=cscx=1sinx

Now As we know the quotient rule of derivative,

d(y1y2)dx=y2dy1dxy1dy2dxy22

So applying this rule, we get

d(1sinx)dx=(sinx)d(1)dx1d(sinx)dx(sinx)2

d(1sinx)dx=1(cosx)(sinx)2

d(1sinx)dx=(cosx)(sinx)1sinx

d(cscx)dx=cotxcscx

Question:11(v) Find the derivative of the following functions: 3cotx+5cscx

Answer:

Given,

f(x)=3cotx+5cscx

As we know the property

d(y1+y2)dx=dy1dx+dy2dx

Applying the property,

d(3cotx+5cscx)dx=d(3cotx)dx+d(5cscx)dx


d(3cotx+5cscx)dx=3d(cosxsinx)dx+d(5cscx)dx

d(3cotx+5cscx)dx=5cscxcotx+3d(cosxsinx)dx

Now As we know the quotient rule of derivative,

d(y1y2)dx=y2dy1dxy1dy2dxy22

So applying this rule, we get

d(3cotx+5cscx)dx=5cscxcotx+3[sinxd(cosx)dxcosx(d(sinx)dx)sin2x]

d(3cotx+5cscx)dx=5cscxcotx+3[sinx(sinx)cosx(cosx)sin2x]

d(3cotx+5cscx)dx=5cscxcotx+3[sin2xcos2xsin2x]

d(3cotx+5cscx)dx=5cscxcotx3[sin2x+cos2xsin2x]

d(3cotx+5cscx)dx=5cscxcotx3[1sin2x]

d(3cotx+5cscx)dx=5cscxcotx3csc2x

Question:11(vi) Find the derivative of the following functions: 5sinx6cosx+7

Answer:

Given,

f(x)=5sinx6cosx+7

Now as we know the property

d(y1+y2)dx=dy1dx+dy2dx

So, applying the property,

f(x)=5cosx6(sinx)+0

f(x)=5cosx+6(sinx)

f(x)=5cosx+6sinx

Question:11(vii) Find the derivative of the following functions: 2tanx7secx

Answer:

Given

f(x)=2tanx7secx

As we know the property

d(y1+y2)dx=dy1dx+dy2dx

Applying this property,

d(2tanx+7secx)dx=2dtanxdx+7dsecxdx

d(2tanx+7secx)dx=2sec2x+7(secxtanx)

d(2tanx+7secx)dx=2sec2x7secxtanx

Limits and derivatives NCERT solutions - Miscellaneous Exercise

Question:1(i) Find the derivative of the following functions from first principle: -x

Answer:

Given.

f(x)=-x

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0(x+h)(x)h

f(x)=limh0hh

f(x)=1

Question:1(ii) Find the derivative of the following functions from first principle: (x)1

Answer:

Given.

f(x)= (x)1

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0(x+h)1(x)1h

f(x)=limh01x+h+1xh

f(x)=limh0x+x+hx(x+h)h

f(x)=limh0h(x+h)(x)h

f(x)=limh01x(x+h)

f(x)=1x2

Question:1(iii) Find the derivative of the following functions from first principle: sin(x+1)

Answer:

Given.

f(x)=sin(x+1)

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0sin(x+h+1)sin(x+1)h

f(x)=limh02cos(x+h+1+x+12)sin(x+h+1x12)h

f(x)=limh02cos(2x+h+22)sin(h2)h

f(x)=limh0cos(2x+h+22)sin(h2)h2

f(x)=cos(2x+0+22)×1

f(x)=cos(x+1)

Question:1(iv) Find the derivative of the following functions from first principle: cos(xπ/8)

Answer:

Given.

f(x)=cos(xπ/8)

Now, As we know, The derivative of any function at x is

f(x)=limh0f(x+h)f(x)h

f(x)=limh0cos(x+hπ/8)cos(xπ/8)h

f(x)=limh02sin(x+hπ/8+xπ/82)sin(x+hπ/8x+π/82)h

f(x)=limh02sin(2x+hπ/42)sin(h2)h

f(x)=limh0sin(2x+hπ/42)sin(h2)h2

f(x)=sin(2x+0π/42)×1

f(x)=sin(xπ/8)

Question:2 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x+a )

Answer:

Given

f(x)= x + a

As we know, the property,

f(xn)=nxn1

applying that property we get

f(x)=1+0

f(x)=1

Question:3 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (px+q)(rx+s)

Answer:

Given

f(x)=(px+q)(rx+s)

f(x)=pr+psx+qrx+qs

As we know, the property,

f(xn)=nxn1

applying that property we get

f(x)=0+ps+qrx2+0

f(x)=ps+q(rx2)

f(x)=ps(qrx2)

Question:4 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax+b)(cx+d)2

Answer:

Given,

f(x)=(ax+b)(cx+d)2

f(x)=(ax+b)(c2x2+2cdx+d2)

f(x)=ac2x3+2acdx2+ad2x+bc2x2+2bcdx+bd2

Now,

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=3ac2x2+4acdx+ad2+2bc2x+2bcd+0

f(x)=3ac2x2+4acdx+ad2+2bc2x+2bcd

f(x)=3ac2x2+(4acd+2bc2)x+ad2+2bcd

Question:5 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): ax+b/cx+d

Answer:

Given,

f(x)=ax+bcx+d

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence, The derivative of f(x) is

d(ax+bcx+d)dx=(cx+d)d(d(ax+b)dx)(ax+b)(d(cx+d)dx)(cx+d)2

d(ax+bcx+d)dx=(cx+d)a(ax+b)c(cx+d)2

d(ax+bcx+d)dx=acx+adacxbc(cx+d)2

d(ax+bcx+d)dx=adbc(cx+d)2

Hence Derivative of the function is

adbc(cx+d)2 .

Question:6 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

1+1x11x

Answer:

Given,

f(x)=1+1x11x

Also can be written as

f(x)=x+1x1

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence, The derivative of f(x) is

d(x+1x1)dx=(x1)d(d(x+1)dx)(x+1)(d(x1)dx)(x1)2

d(x+1x1)dx=(x1)1(x+1)1(x1)2

d(x+1x1)dx=x1x1(x1)2

d(x+1x1)dx=2(x1)2

Hence Derivative of the function is

2(x1)2

Question:7 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

1ax2+bx+c

Answer:

Given,

f(x)=1ax2+bx+c

Now, As we know the derivative of any such function is given by

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence, The derivative of f(x) is

d(1ax2+bx+c)dx=(ax2+bx+c)d(d(1)dx)1(d(ax2+bx+c)dx)(ax2+bx+c)2

d(1ax2+bx+c)dx=0(2ax+b)(ax2+bx+c)2

d(1ax2+bx+c)dx=(2ax+b)(ax2+bx+c)2

Question:8 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

ax+bpx2+qx+r

Answer:

Given,

f(x)=ax+bpx2+qx+r

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence, The derivative of f(x) is

d(ax+bpx2+qx+r)dx=(px2+qx+r)d(d(ax+b)dx)(ax+b)(d(px2+qx+r)dx)(px2+qx+r)2

d(ax+bpx2+qx+r)dx=(px2+qx+r)a(ax+b)(2px+q)(px2+qx+r)2

d(ax+bpx2+qx+r)dx=apx2+aqx+ar2apx2aqx2bpxbq(px2+qx+r)2

d(ax+bpx2+qx+r)dx=apx2+ar2bpxbq(px2+qx+r)2

Question:9 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

px2+qx+rax+b

Answer:

Given,

f(x)=px2+qx+rax+b

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence, The derivative of f(x) is

d(px2+qx+rax+b)dx=(ax+b)d(d(px2+qx+r)dx)(px2+qx+r)(d(ax+b)dx)(ax+b)2

d(px2+qx+rax+b)dx=(ax+b)(2px+q)(px2+qx+r)(a)(ax+b)2

d(px2+qx+rax+b)dx=2apx2+aqx+2bpx+bqapx2aqxar(ax+b)2

d(px2+qx+rax+b)dx=apx2+2bpx+bqar(ax+b)2

Question:10 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

ax4bx2+cosx

Answer:

Given

f(x)=ax4bx2+cosx

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=4ax5(2bx3)+(sinx)

f(x)=4ax5+2bx3sinx

Question:11 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): 4x2

Answer:

Given

f(x)=4x2

It can also be written as

f(x)=4x122

Now,

As we know, the property,

f(xn)=nxn1

and the property

d(y1+y2)dx=dy1dx+dy2dx

applying that property we get

f(x)=4(12)x120

f(x)=2x12

f(x)=2x

Question:12 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

(ax+b)n

Answer:

Given

f(x)=(ax+b)n

Now, As we know the chain rule of derivative,

[f(g(x))]=f(g(x))×g(x)

And, the property,

f(xn)=nxn1

Also the property

d(y1+y2)dx=dy1dx+dy2dx

applying those properties we get,

f(x)=n(ax+b)n1×a

f(x)=an(ax+b)n1

Question:13 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax+b)n(cx+d)m

Answer:

Given

f(x)=(ax+b)n(cx+d)m

Now, As we know the chain rule of derivative,

[f(g(x))]=f(g(x))×g(x)

And the Multiplication property of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

And, the property,

f(xn)=nxn1

Also the property

d(y1+y2)dx=dy1dx+dy2dx

Applying those properties we get,

f(x)=(ax+b)n(m(cx+d)m1)+(cx+d)(n(ax+b)n1)

f(x)=m(ax+b)n(cx+d)m1+n(cx+d)(ax+b)n1

Question:14 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): sin(x+a)

Answer:

Given,

f(x)=sin(x+a)

Now, As we know the chain rule of derivative,

[f(g(x))]=f(g(x))×g(x)

Applying this property we get,

f(x)=cos(x+a)×1

f(x)=cos(x+a)

Question:15 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): cscxcotx

Answer:

Given,

f(x)=cscxcotx

the Multiplication property of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

Applying the property

d(cscx)(cotx))dx=cscxdcotxdx+cotxdcscxdx

d(cscx)(cotx))dx=cscx(csc2x)+cotx(cscxcotx)

d(cscx)(cotx))dx=csc3xcot2xcscx

Hence derivative of the function is csc3xcot2xcscx .

Question:16 Find the derivative of the following functions (it is to be understood that a, b, c, d,p, q, r and s are fixed non-zero constants and m and n are integers):

cosx1+sinx

Answer:

Given,

f(x)=cosx1+sinx

Now, As we know the derivative of any function

d(cosx1+sinx)dx=(1+sinx)d(dcosxdx)cosx(d(1+sinx)dx)(1+sinx)2

Hence, The derivative of f(x) is

d(cosx1+sinx)dx=(1+sinx)(sinx)cosx(cosx)(1+sinx)2

d(cosx1+sinx)dx=sinxsin2xcos2x(1+sinx)2

d(cosx1+sinx)dx=sinx1(1+sinx)2

d(cosx1+sinx)dx=1(1+sinx)

Question:17 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): sinx+cosxsinxcosx

Answer:

Given

f(x)=sinx+cosxsinxcosx

Also can be written as

f(x)=tanx+1tanx1

which further can be written as

f(x)=tanx+tan(π/4)1tan(π/4)tanx

f(x)=tan(xπ/4)

Now,

f(x)=sec2(xπ/4)

f(x)=1cos2(xπ/4)

Question:18 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

secx1secx+1

Answer:

Given,

f(x)=secx1secx+1

which also can be written as

f(x)=1cosx1+cosx

Now,

As we know the derivative of such function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

So, The derivative of the function is,

d(1cosx1+cosx)dx=(1+cosx)d(d(1cosx)dx)(1cosx)(d(1+cosx)dx)(1+cosx)2

d(1cosx1+cosx)dx=(1+cosx)((sinx))(1cosx)(sinx)(1+cosx)2

d(1cosx1+cosx)dx=sinx+sinxcosx+sinxcosxsinx(1+cosx)2

d(1cosx1+cosx)dx=2sinx(1+cosx)2

Which can also be written as

d(1cosx1+cosx)dx=2secxtanx(1+secx)2 .

Question:19 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): sinnx

Answer:

Given,

f(x)=sinnx

Now, As we know the chain rule of derivative,

[f(g(x))]=f(g(x))×g(x)

And, the property,

f(xn)=nxn1

Applying those properties, we get

f(x)=nsinn1xcosx

Hence Derivative of the given function is nsinn1xcosx

Question:20 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

a+bsinxc+dcosx

Answer:

Given Function

f(x)=a+bsinxc+dcosx

Now, As we know the derivative of any function of this type is:

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence derivative of the given function will be:

d(a+bsinxc+dcosx)dx=(c+dcosx)(d(a+bsinx)dx)(a+bsinx)(d(c+dcosxx)dx)(c+dcosx)2

d(a+bsinxc+dcosx)dx=(c+dcosx)(bcosx)(a+bsinx)(d(sinx))(c+dcosx)2

d(a+bsinxc+dcosx)dx=cbcosx+bdcos2x+adsinx+bdsin2x(c+dcosx)2

d(a+bsinxc+dcosx)dx=cbcosx+adsinx+bd(sin2x+cos2x)(c+dcosx)2

d(a+bsinxc+dcosx)dx=cbcosx+adsinx+bd(c+dcosx)2

Question:21 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

sin(x+a)cosx

Answer:

Given,

f(x)=sin(x+a)cosx

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence the derivative of the given function is:

d(sin(x+a)cosx)dx=(cosx)(d(sin(x+a))dx)sin(x+a)(d(cosx)dx)(cosx)2

d(sin(x+a)cosx)dx=(cosx)(cos(x+a))sin(x+a)(sin(x))(cosx)2

d(sin(x+a)cosx)dx=(cosx)(cos(x+a))+sin(x+a)(sin(x))(cosx)2

d(sin(x+a)cosx)dx=cos(x+ax)(cosx)2

d(sin(x+a)cosx)dx=cos(a)(cosx)2

Question:22 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): x4(5sinx3cosx)

Answer:

Given

f(x)=x4(5sinx3cosx)

Now, As we know, the Multiplication property of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

Hence derivative of the given function is:

d(x4(5sinx3cosx))dx=x4d(5sinx3cosx)dx+(5sinx3cosx)dx4dx

d(x4(5sinx3cosx))dx=x4(5cosx+3sinx)+(5sinx3cosx)4x3

d(x4(5sinx3cosx))dx=5x4cosx+3x4sinx+20x3sinx12x3cosx

Question:23 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x2+1)cosx

Answer:

Given

f(x)=(x2+1)cosx

Now, As we know the product rule of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

The derivative of the given function is

d((x2+1)cosx)dx=(x2+1)dcosxdx+cosxd(x2+1)dx

d((x2+1)cosx)dx=(x2+1)(sinx)+cosx(2x)

d((x2+1)cosx)dx=x2sinxsinx+2xcosx

Question:24 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax2+sinx)(p+qcosx)

Answer:

Given,

f(x)=(ax2+sinx)(p+qcosx)

Now As we know the Multiplication property of derivative,(the product rule)

d(y1y2)dx=y1dy2dx+y2dy1dx

And also the property

d(y1+y2)dx=dy1dx+dy2dx

Applying those properties we get,

d((ax2+sinx)(p+qcosx))dx=(ax2+sinx)d(p+qcosx)dx+(p+qx)d(ax2+sinx)dx

d((ax2+sinx)(p+qcosx))dx=(ax2+sinx)(qsinx)+(p+qx)(2ax+cosx)

f(x)=aqx2sinxqsin2x+2apx+pcosx+2aqx2+qxcosx

f(x)=x2(aqsinx+2aq)+x(2ap+qcosx)+pcosxqsin2x

Question:25 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x+cosx)(xtanx)

Answer:

Given,

f(x)=(x+cosx)(xtanx)

And the Multiplication property of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

Also the property

d(y1+y2)dx=dy1dx+dy2dx

Applying those properties we get,

d((x+cosx)(xtanx))dx=(x+cosx)d(xtanx)dx+(xtanx)d(x+cosx)dx

=(x+cosx)(1sec2x)+(xtanx)(1sinx)

=(x+cosx)(tan2x)+(xtanx)(1sinx)

=(tan2x)(x+cosx)+(xtanx)(1sinx)

Question:26 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

4x+5sinx3x+7cosx

Answer:

Given,

f(x)=4x+5sinx3x+7cosx

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Also the property

d(y1+y2)dx=dy1dx+dy2dx

Applying those properties,we get

d(4x+5sinx3x+7cosx)dx=(3x+7cosx)d(d(4x+5sinx)dx)(4x+5sinx)(d(3x+7cosx)dx)(3x+7cosx)2

d(4x+5sinx3x+7cosx)dx=(3x+7cosx)(4+5cosx)(4x+5sinx)(37sinx)(3x+7cosx)2

=12x+28cosx+15xcosx+35cos2x12x15sinx+28xsinx+35sin2x(3x+7cosx)2

=12x+28cosx+15xcosx12x15sinx+28xsinx+35(sin2x+cos2x)(3x+7cosx)2

=28cosx+15xcosx15sinx+28xsinx+35(3x+7cosx)2

Question:27 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

x2cos(π/4)sinx

Answer:

Given,

f(x)=x2cos(π/4)sinx

Now, As we know the derivative of any function

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Hence the derivative of the given function is

d(x2cos(π/4)sinx)dx=(sinx)d(d(x2cos(π/4))dx)(x2cos(π/4))(dsinxdx)sin2x

d(x2cos(π/4)sinx)dx=(sinx)(2xcos(π/4))(x2cos(π/4))(cosx)sin2x

d(x2cos(π/4)sinx)dx=2xsinxcos(π/4)x2cosxcos(π/4)sin2x

Question:28 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

x1+tanx

Answer:

Given

f(x)=x1+tanx

Now, As we know the derivative of any function

d(x1+tanx)dx=(1+tanx)d(dxdx)x(d(1+tanx)dx)(1+tanx)2

d(x1+tanx)dx=(1+tanx)(1)x(sec2x)(1+tanx)2

d(x1+tanx)dx=1+tanxxsec2x(1+tanx)2

Question:29 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x+secx)(xtanx)

Answer:

Given

f(x)=(x+secx)(xtanx)

Now, As we know the Multiplication property of derivative,

d(y1y2)dx=y1dy2dx+y2dy1dx

Also the property

d(y1+y2)dx=dy1dx+dy2dx

Applying those properties we get,

the derivative of the given function is,

d((x+secx)(xtanx)dx=(x+secx)d(xtanx)dx+(xtanx)d(x+secx)dx

d((x+secx)(xtanx)dx=(x+secx)(1sec2x)+(xtanx)(1+secxtanx)

Question:30 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

xsinnx

Answer:

Given,

f(x)=xsinnx

Now, As we know the derivative of any function

d(y1y2)dx=y2d(dy1dx)y1(dy2dx)y22

Also chain rule of derivative,

[f(g(x))]=f(g(x))×g(x)

Hence the derivative of the given function is

d(xsinnx)dx=sinnxd(dxdx)x(d(sinnx)dx)sin2nx

d(xsinnx)dx=sinnx(1)x(nsinn1x×cosx)sin2nx

d(xsinnx)dx=sinnxxcosxnsinn1xsin2nx

d(xsinnx)dx=sinnxxcosxnsinn1xsin2nx

d(xsinnx)dx=sinn1x(sinxnxcosx)sin2nx

d(xsinnx)dx=(sinxnxcosx)sinn+1x

limits and derivatives ncert solutions - Topics

13.1 Introduction

13.2 Intuitive Idea of Derivatives

13.3 Limits

13.4 Limits of Trigonometric Functions

13.5 Derivatives

Interested students can practice class 11 maths ch 13 question answer using following exercises.

Limit of a function(f) at a point is the common value of the left-hand limits and right-hand limits if they coincide.

For a function 'f' and a real number 'a'. limxaf(x) and f(a) may not be the same (One may be defined and not the other one).

For functions 'f' and 'g' the following properties holds:

limxa[f(x)±g(x)]=limxaf(x)±limxag(x)limxa[f(x).g(x)]=limxaf(x).limxag(x)limxa[f(x)g(x)]=limxaf(x)limxag(x)

  • Some of the standard limits are-

limxaxnanxa=nan1limxasinxx=1limxa1cosxx=0

  • The derivative of a function 'f' at 'a' is defined by-

f=limh0f(a+h)f(a)h

  • The derivative of a function 'f' at any point 'x' is defined by-

f(x)=df(x)dx=limh0f(x+h)f(x)h

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Key Features Of Limits And Derivatives Class 11 Solutions

Comprehensive coverage of concepts: The ch 13 maths class 11 provides a thorough and comprehensive coverage of the fundamental concepts of limits and derivatives, which are essential for understanding the principles of calculus.

Simple and easy language: The class 11 maths limits and derivatives is written in a simple and easy-to-understand language, making it accessible to students of all levels of mathematical aptitude.

Step-by-step explanations: The chapter provides step-by-step explanations of the concepts, with worked-out examples and exercises that help students to master the topics covered in the chapter.

NCERT Solutions for Class 11- Subject Wise

Tip- Limits class 11 is more on the theory-based so you should understand every theorem and try to solve questions based on the theorem. Derivatives required more practice of problems. There are 30 questions in the miscellaneous exercise. You should solve miscellaneous exercises also to get command on this chapter with the help of NCERT solutions for class 11 maths chapter 13 limits and derivatives.

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. What is the effective way to study limits and derivatives class 11 NCERT solutions?

Effective way to study chapter 13 class 11 maths includes following points:

  1. Begin by familiarizing yourself with the basic concepts covered in the chapter.
  2. Understand the fundamental topics of calculus, including both differential and integral calculus.
  3. Memorize the essential formulas that are used in the chapter.
  4. Focus on learning about limits and their properties in detail.
  5. Master the fundamental theorem of calculus and its significance.
  6. To solidify your understanding of the concepts, practice problems on a regular basis.
2. What are the topics included in limits and derivatives class 11 solutions?

class11 maths ch13 includes following topics:

1. Introduction
2. Intuitive Idea of Derivatives
3. Limits
4. Limits of Trigonometric Functions
5. Derivatives

3. How does the NCERT solutions for class 11 chapter 13 maths are helpful ?

NCERT solutions for class 11 math ch 13 are helpful to the students in solving the NCERT problems as well building foundation for class 12th chapters such as differentiation, integration and many. These solutions are provided in a detailed manner which can be understood by an average student also.

4. Where can I find the complete class 11 maths NCERT solutions chapter 13?

Students can get the detailed NCERT solutions for class 11 maths here. they can practice these solutions to build concepts that lead confidence during exam and ultimately help to score well in the exam.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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