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RD Sharma Class 12 Exercise 10.1 Differentiation Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 10.1 Differentiation Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 20, 2022 05:28 PM IST

RD Sharma Class 12th Exercise 10.1 is the most preferred reference material by many CBSE schools as well as the students. The high standard of the solved sums present in this book makes the students understand every concept in-depth. For many years, the questions for the board examinations have been taken from the RD Sharma books. When the students use this book for their practice, they would be exam-ready effortlessly. RD Sharma Solutions
Chapter 10 for class 12 mathematics, Differentiation has 8 exercises in total. The first exercise, ex 10.1 has basic differentiation functions learned in the previous academic year. Using the RD Sharma Class 12th Exercise 10.1 book, the students will be able to solve all the problems easily. The simple tricks and tips to solve the differentiation sums are present in the Class 12th RD Sharma Chapter 10 Exercise 10.1 Solutions material. Concepts like logarithmic differentiation, differentiation of inverse trigonometric functions, and various other differentiating methods are included in this exercise.

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise


Differentiation Excercise: 10.1

Differentiation exercise 10 .1 question 1

Answer:
(ex)
Hint:
Use the first principle to find the differentiation of
(ex)
Given:
(ex)

Solution:
Let
f(x)=ex
f(x+h)=e(x+h)
So, we will use formula of differentiation by first principle,
ddx(f(x))=limh0f(x+h)f(x)h
ddx=limhoe(x+h)exh
=limh0ex×ehexh
=limh0ex(eh1)h
=limh0ex(eh1)h×(1)
=limh0ex×(eh1h)×(1)

=limh0ex(1)×(1) [limh0ex1x]=1
=(ex)
Hence, the differentiation of ex is (ex)

Differentiation exercise 10.1 question 2

Answer:
(3e3x)
Hint:
Use first principle to find the differentiation of (e3x)
Given:
(e3x)
Solution:
Let
f(x)=e3x
f(x+h)=e3(x+h)
So, now we will use formula of differentiation by first principle
ddxf(x)=limh0f(x+h)f(x)h
ddx(e3x)=limh0e3(x+h)e3xh
=limh0e3x×e3he3xh
=limh0e3x(e3h1)h
Multiply and divide by 3
=limh0e3x(e3h1)3h×3
=limh0e3x×1×3 [limh0(ex1x)=1]
=3e3x
Hence, the differentiation of e3x is 3e3x.

Differentiation exercise 10.1 question 3

Answer:
ae(ax+b)
Hint:
Use first principle to find the differentiation
Given:
(eax+b)
Solution:
Let

f(x)=eax+b
f(x+h)=ea(x+h)+b
Now, we will use formula of differentiation by first principle
ddx(f(x))=limh0f(x+h)f(x)h
ddx(eax+b)=limh0ea(x+h)+beax+bh
=limh0eax+ah+beax+bh
=limh0(eax+b)(eah)eax+bh
=limh0eax+b(eah1)h
Multiply and divide by a
=limh0eax+b(eah1)ah×a
=limh0(eax+b)×a [limh0ex1x=1]
=aeax+b
Hence, the differentiation of eax+b is aeax+b.

Differentiation exercise 10.1 question 4

Answer:
(sinx)ecosx
Hint:
Use first principle to find the differentiation
Given:
(ecosx)
Solution:
Let
f(x)=ecosx
f(x+h)=ecos(x+h)
Now, we will use formula of differentiation by first principle
ddx(f(x))=limh0f(x+h)f(x)h
ddx(ecosx)=limh0ecos(x+h)ecosxh
Take ecosx common from the numerator
=limh0(ecosx)[ecos(x+h)ecosx1]h
=limh0ecosx[ecosx(x+h)cosx1]h [aman=amn]

Multiply and divide [cos(x+h)cosx]
=limh0ecosx[ecos(x+h)cosx1](cos(x+h)cosx)×h×[cos(x+h)cosx]

=limh0ecosx×(cos(x+h)cosxh) [limh0ex1x=1]

=limh0ecosx×(cos(x+h)cosxh) [cosAcosB=2sinA+B2sinAB2]
=limh0ecosx×[2sinx+h+x2×sinx+hx2h]
ecosx Is a function of f(x) and limit is applied on variable ‘h’
So,ecosx can be taken outside
=ecosx×limh02sin(2x+h2 )×sin(h2)h
Multiply the denominator by 2 and divide the denominator by 2
=ecosx×limh02sin(2x+h2)2×sin(h2)(h2)
=ecosx×limh0[sin(2x+h2)]×1 [limh0sinxx=1]
=ecosx×(sinx)=sinxecosx
Hence, the differentiation of ecosx is [sinxecosx]


Differentiability exercise 10.1 question 5

Answer:
e2x2x
Hint:
Use first principle formula to find the differentiation
Given:
e2x
Solution:
Let,
f(x)=e2x
f(x+h)=e2(x+h)
Now, we will use the formula of first principle to find the differentiation
ddx(f(x))=limh0f(x+h)f(x)h
ddx(e2x)=limh0e2(x+h)e2xh
Take 2e2x common from numerator
=limh0e2x[(e2(x+h)e2x)1]h
=limh0e2x[([e2(x+h)])1]h [aman=amn]

Multiply and divide by (2(x+h)2x)
=limh0e2x×[(e2(x+h)2x)1](2(x+h)2x)×[2(x+h)2xh]

=limh0e2x×1×[2(x+h)2xh] [limh0ex1x=1]

=limh0e2x×[2(x+h)2xh]
Now, e2x is a function of x and unit applied on variable ‘h’. So,e2x can be taken outside became w.r.t variable ‘h’, x will remain constant
=e2x×limh0[2(x+h)2xh]

Rationalising the numerator
=e2x×limh0[2(x+h)2xh×2(x+h)+2x2(x+h)+2x]
=e2x×limh0[(2(x+h))2(2x)2h×(2(x+h))] [(a+b)(ab)=a2b2]
=e2x×limh02(x+h)2xh(2(x+h)+2x)
=e2x×limh02x+2h2xh(2(x+h)+2x)
=e2x×limh02hh(2(x+h)+2x)
=e2x×limh02h(2(x+h)+2x)
=e2x×222x
=e2x2x
Hence, the differentiation of e2x ise2x2x

Differentiation exercise 10.1 question 6

Answer:
(tanx)
Hint:
Use first principle formula to find the differentiation
Given:
log(cosx)
Solution:
Let,
fx=log(cosx)
f(x+h)=log(cos(x+h))
Now, we will use formula of first principle to find the differentiation
ddx(f(x))=limh0f(x+h)f(x)h
ddx(log(cosx))=limh0log(cos(x+h))log(cosx)h
=limh0log(cos(x+h)cosx)h [logAlogB=logAB]
Add and subtract 1 in the argument of log in numerator

=limh0log[1+cos(x+h)cosx1]h
=limh0log[1+cos(x+h)cosxcosx]h
Multiply and divide by =cos(x+h)cosxh
=limh0log[1+cos(x+h)cosxcosx]h×[cos(x+h)cosxcosx]×[cos(x+h)cosxcosx]
=limh0(1h)×log(1+cos(x+h)cosxcosx)cos(x+h)cosxcosx×cos(x+h)cosxcosx
=limh0(1h)×1×cos(x+h)cosxcosx [limh0log(1+x)x=1]
=limh0cos(x+h)cosxh×cosx

=limh02sin((x+h)+x2)sin((x+h)x2)h×cosx [cosAcosB=2sin(A+B2)sin(AB2)]
=limh02sin(2x+h2)sin(h2)hcosx
Multiply the denominator and divide the denominator by 2
=limh02(2x+h2)cosxsin(h2)(h2)×2 [limh0sinxx=1]
=limh0sin(2x+h2)cosx×1
=sin(2x2)cosx
=sinxcosx
=tanx
Hence, the differentiation of log(cosx)=(tanx)


Differentiation exercise 10.1 question 7

Answer:
ecotx×cosec2x2cotx
Hint:
Use first principle formula to find the differentiation
Given:
ecotx
Solution:
Let,
f(x)=ecotxf(x+h)=ecot(x+h)
Now, we will use the formula of first principle to find the differentiation
ddx(f(x))=limh0f(x+h)f(x)hddx(ecotx)=limh0ecot(x+h)ecotxh
Take ecotx common from numerator
=limh0ecotx(ecot(x+h)ecotx1)h
=limh0ecotx(ecot(x+h)cotx1)h [aman=amn]
ecotx Is a function of variable x and limit is applied on variable ‘h’
So,
=ecotx×limh0e(cot(x+h)cotx)1h
Multiply and divide by (cot(x+h)cotx)
=ecotx×limh0[ecot(x+h)cotx1](cot(x+h)cotx)×[cot(x+h)cotxh]
=ecotx×limh0(1)×[cot(x+h)cotxh] [limh0ex1x=1]
Rationalising the numerator
=ecotx×limh0[cot(x+h)cotxh×cot(x+h)+cotxcot(x+h)+cotx]
=ecotx×limh0cot(x+h)cotxh(cot(x+h)+cotx) [(a+b)(ab)=a2b2][(a+b)(ab)=a2b2]

=ecotx×limh0cot(x+h)×cotx+1cot(x+hx)h(cot(x+h)+cotx)[cot(A+B)=cotAcotB+1cotAcotB]

=(ecotx)×limh0cot(x+h)cotx+1h(coth)(cot(x+h)+cotx)

=ecotx×limh0cot(x+h)cotx+1(htanh)(cot(x+h)+cotx)[tanθ=1cotθ]


=ecotx×limh0tanhh×[cot(x+h)cotx+1cot(x+h)+cotx]
=ecotx×1×limh0cot(x+h)cotx+1cot(x+h)+cotx [limx0tanxx=1]
=ecotx×cot2x+12cotx
=ecotx×cosec2x2cotx [1+cot2θ=cosec2θ]
ddx(ecotx)=ecotx×cosec2x2cotx
Hence, the differentiation of ecotx is ecotx×cosec2x2cotx

Differentiation exercise 10.1 question 8

Answer:
ex2(x2+2x)
Hint:
Use first principle formula to find the differentiation
Given:
x2ex
Solution:
Let,f(x)=x2exf(x+h)=(x+h)2e(x+h)

Now, we will use the formula of first principle

ddx(f(x))=limh0f(x+h)f(x)hddx(x2ex)=limh0(x+h)2e(x+h)x2exh

=limh0(x2+2hx+h2)ex+hx2exh [(a+b)2=a2+2ab+b2]


=limh0[(x2ex+hx2ex)h+2hxex+hh+h2ex+hh]=limh0[x2exehx2exh+2xex+h+hex+h]=limh0[x2ex(eh1)h+2xex+h+hex+h]=limh0[x2ex(eh1h)+2xex+h+hex+h]

=limh0[(x2ex)×1+2xex+h+hex+h] [limh0ex1x=1]

=x2ex+2xex+(0×ex)=x2ex+2xex=ex(x2+2x)ddx(x2ex)=ex(x2+2x)

Hence, the differentiation of x2ex is ex(x2+2x)

Differentiation exercise 10.1 question 9

Answer:

cotx

Hint:

Use first principle formula to find the differentiation

Given:

log(cosecx)

Solution:

Let,

f(x)=log(cosecx)f(x+h)=log(cosec(x+h))

Now, we will use the formula of first principle

ddx(f(x))=limh0f(x+h)f(x)hddx(log(cosecx))=limh0log(cosec(x+h)log(cosecx))h

=limh0log(cosec(x+h)cosecx)h[logalogb=log(ab)]

=limh0log(1sin(x+h)1sinx)h[cosecx=1sinx]

=limh0log(sinxsin(x+h))h

Add 1 and subtract 1 from argument of the numerator


=limh0log(1+sinxsin(x+h)1)h=limh0log(1+sinxsin(x+h)sin(x+h))h

Multiply and divide by sinxsin(x+h)sin(x+h)

=limh0log(1+sinxsin(x+h)sin(x+h))[sinxsin(x+h)sin(x+h)]×[sinxsin(x+h)sin(x+h)]h

=limh0×sinxsin(x+h)hsin(x+h)[limh0log(1+x)x=1]

=limh02cos(2x+h2)×sin(h2)hsin(x+h) [sinAsinB=2cos(A+B2)sin(AB2)]

=limh02cos(2x+h2)×sin(h2)hsin(x+h)

Divide and multiply the denominator by (-2)

=limh02cos(2x+h2)sin(h2)(2)×sin(x+h)(h2)

=limh0cos(2x+h2)sin(x+h)×(sin(h2)(h2)) [limh0sinxx=1]

=limh0cos(2x+h2)sin(x+h)=cotxddx(logcosecx)=cotx

Hence, the differentiation of cotx

Differentiation exercise 10.1 question 10

Answer:
21(2x+3)2
Hint:
Use first principle formula to find the differentiation
Given:
sin1(2x+3)
Solution:
Let,
f(x)=sin1(2x+3)f(x+h)=sin1(2(x+h)+3)=sin1(2x+2h+3)
Now, we will use formula of first principle to find the differentiation
ddx(f(x))=limh0f(x+h)f(x)hddx(sin1(2x+3))=limh0sin1(2x+2h+3)sin1(2x+3)h=limh0sin1[(2x+2h+3)1(2x+3)2(2x+3)1(2x+2h+3)2]h
[sin1xsin1y=sin1[x1y2y1x2]]
Let,

[(2x+2h+3)1(2x+3)2(2x+3)1(2x+2h+3)2]=z=limh0sin1zh
Multiply and divide by z
=limh0sin1zh×zh
=limh0(1)×zh [limh0sin1xx=1]
=limh0zh
Put the value of z now
=limh0(2x+2h+3)1(2x+3)2(2x+3)1(2x+2h+3)2h
Multiply and divide by (2x+2h+3)1(2x+3)2(2x+3)1(2x+2h+3)2
[(2x+2h+3)1(2x+3)2(2x+3)1(2x+2h+3)2]=limh0×[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]=limh0((2x+2h+3)1(2x+3)2)2((2x+3)1(2x+2h+3)2)2h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]
(a+b)(ab)=a2b2
=limh0[(2x+2h+3)2(1(2x+3)2)][(2x+3)2(1(2x+2h+3)2)]h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]=limh0[(2x+3)+2h]2[1(2x+3)2][(2x+3)2][1((2x+3)+2h)2]h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]
=limh0{[(2x+3)2+4h2+4h(2x+3)][1(2x+3)2]}{(2x+3)2(1(2x+3)24h24h(2x+3))}h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]
((a2+b2)=a2+b2+2ab)
=limh0[(2x+3)2+4h2+4h(2x+3)(2x+3)44h2(2x+3)24h(2x+3)3(2x+3)2+(2x+3)4+4h2(2x+3)2+4h(2x+3)3]h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]
=limh04h2+4h(2x+3)h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]=limh04h(h+(2x+3))h[(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2]=limh04(h+(2x+3))(2x+2h+3)1(2x+3)2+(2x+3)1(2x+2h+3)2
=4(2x+3)(2x+3)1(2x+3)2+(2x+3)1(2x+3)2=4(2x+3)2×(2x+3)1(2x+3)2=21(2x+3)2ddx(sin1(2x+3))=21(2x+3)2
Hence, the differentiation of sin1(2x+3) Is 21(2x+3)2.

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RD Sharma Chapter-wise Solutions

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