Careers360 Logo
School
Search Colleges, Exams, Schools & more
RD Sharma Class 12 Exercise 10.3 Differentiation Solutions Maths-Download PDF Free Online

RD Sharma Class 12 Exercise 10.3 Differentiation Solutions Maths-Download PDF Free Online

Updated on Jan 20, 2022 05:22 PM IST

The RD Sharma solution books are the top-rated reference materials used by the students preparing for their public exams. It lends a helping hand in every aspect of doubts that a student might encounter while studying. When the students continuously practice the sums given in this book, mathematics would be simple and easy for them.
Rd Sharma Class 12th Exercise 10.3 has solved every problem of a student regarding Differentiation. This particular exercise has 49 questions, including subparts, that are formatted by a team of experts. The questions will help every student solve the queries regarding differentiating the functions w.r.t to x, Differentiation of inverse trigonometric functions, recapitulation products, Differentiation of constant values are included in Rd Sharma Class 12th Exercise 10.3.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
  2. Differentiation Excercise: 10.3
  3. RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.3

Differentiation exercise 10.3 question 1

Answer : dydx=21−x2
Hint:
$cos−1⁡(cos⁡θ)=θ  if  θε[0,π]cos−1⁡(cos⁡θ)=−θ  if  θε[−π,0]$
Given:
cos−1⁡{2x1−x2},12<x<1
Solution:
$Letsy=cos−1⁡{2x1−x2}Letsx=cos⁡θ⇒θ=cos−1⁡xNow  y=cos−1⁡{2cos⁡θ1−cos2⁡θ}$
$Using∴cos2⁡θ+sin2⁡θ=1,2sin⁡θcos⁡θ=sin⁡2θy=cos−1⁡{2cos⁡θsin2⁡θ}$
$y=cos−1⁡(2cos⁡θsin⁡θ]y=cos−1⁡(sin⁡2θ)∴cos⁡(π2−θ)⇒sin⁡θy=cos−1⁡(cos⁡(π2−2θ))$
Now by considering the limits,
$12<x<1⇒12<cos⁡θ<1⇒0<θ<π4          ∴[cos⁡90∘=0,cos⁡0∘=1]⇒0<2θ<π2⇒0>−2θ>−π2⇒π2>π2−2θ>π2−π2$
Therefore,
$y=cos−1⁡(cos⁡(π2−2θ))y=cos−1⁡(cos⁡(π2−2θ))y=(π2−2θ)              $ {cos−1⁡(cos⁡θ)=θ}
y=π2−2cos−1⁡x
 if θε[0,π]
Differentiating with Respect to x, we get
$dydx=ddx(π2−2cos−1⁡x)∴ddx(cos−1⁡x)⇒−11−x2∴ddx (constant) =0dydx=0−2(−11−x2)dydx=21−x2$

Differentiation exercise 10.3 question 2

Answer : dydx=12(−11−x2)
Hint:
$cos−1⁡(cos⁡θ)=θif  θε[0,π]cos−1⁡(cos⁡θ)=−θif   θε[−π,0]$
Given:
$cos−1⁡{1+x2},−1<x<1$
Solution:
$Let,y=cos−1⁡{1+x2}Letx=cos⁡2θ,θ=12cos−1⁡xNow,y=cos−1⁡{1+cos⁡2θ2}$
$By using cos⁡2θ=2cos2⁡θ−1y=cos−1⁡{2cos2⁡θ2}y=cos−1⁡(cos⁡θ)∴cos−1⁡(cos⁡θ)=θ$
$Nowy=cos−1⁡(cos⁡θ)y=θy=12cos−1⁡x       ∴cos−1⁡(cos⁡θ)=θifθε[0,π]$
Differentiating with respect to x, we get
$dydx⇒12(−11−x2)$
As we know,
$ddx (constants)=0ddx(cos−1⁡x)=−11−x2$
Hence, final answer is $12(−11−x2)$

Differentiation exercise 10.3 question 3

Answer : -−121−x2
Hint:
Use substitution method to differentiate this function
Given:
sin−1⁡{1−x2}},0<x<1
Solution:
$Let⁡y=sin−1⁡{1−x2}$
$ let x=cos⁡2θ[θ=12cos−1⁡x]  −(1)$
$∴y=sin−1⁡{1−cos⁡2θ2}=sin−1⁡{2sin2⁡θ2}y=sin−1⁡{sin2⁡θ}    [∵1−cos⁡2θ=2sin2⁡θ]=sin−1⁡(sin⁡θ)y=θ=12cos−1⁡x[ using (i)] $
Now differentiating y w.r.t x then

$dydx=12d(cos−1⁡x)dx=12⋅−11−x2=−121−x2[∵ddxcos−1⁡x=−11−x2]∴dydx=−121−x2$

Differentiation exercise 10.3 question 4

Answer :
dydx=−11−x2
Hint:
ddx(xn)=nxn−1;ddx(constant)=0
Given:
sin−1⁡{1−x2},0<x<1
Solution:
$\begin {array}{l} Let \ \ x=\cos \theta\\ \theta=\cos ^{-1} x\\ y=\sin ^{-1}\left\{\sqrt{ \left.1-\cos ^{2} \theta\right\}}\right.\\ \therefore \cos ^{2} \theta+\sin ^{2} \theta=1\\ y=\sin ^{-1}\left\{\sqrt{\sin ^{2} \theta}\right\}\\ \mathrm{y}=\sin ^{-1}\{\sin \theta\} \end{}$
Considering the limits,
$\begin {array}{l} 0<\cos \theta<1 \\ 0<\theta<\frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ \left\{\cos \frac{1}{2}=1\right\}\\ Now, y=\sin ^{-1}(\sin \theta)\\ \mathrm{y}=\theta\\ \left\{\sin ^{-1}(\sin \theta)=\theta\right\}\\ if \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \\ y=\cos ^{-1} x \end{}$
differentiating with respects to x, we get
$\begin {array}{l} \frac{d y}{d x}=\frac{\mathrm{d} (\cos^{-1}x)}{\mathrm{d} x}\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right) \Rightarrow \frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\\ \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \end{}$

Differentiation exercise 10.3 question 6


Answer :dydx=aa2+x2
Hint:

Given:
sin−1⁡{xx2+a2}
Solution:
$Let y=sin−1⁡{−xx2+a2}Let x=atan⁡θNow,y=sin−1⁡{atan⁡θa2tan2⁡θ+a2}y=sin−1⁡{ atsan θa2(tan2⁡θ+1)} y=sin−1⁡{atan⁡θa(tan2⁡θ+1)}y=sin−1⁡∫atan⁡θasec2⁡θ}$
$ Using :{tan⁡θ=sin⁡θcos⁡θ,cos⁡θ=1sec⁡θ}y=sin−1⁡{sin⁡θ}y=θ         ∴{sin−1⁡θ(sin⁡θ)=θ} if θε[−π2,π2]x=atan⁡θx=atan⁡θθ=tan−1⁡(xa)y=tan−1⁡(xa)$
differentiating with respects to x, we get
$dydx=ddx(tan−1⁡(xa))∂∂x(tan−1⁡x)=11+x2dydx=(11+(xa)2)×1a$
$dydx=(11+x2a2)×1adydx=(1a2+x2a2)×1adydx=a2a2+x2×1adydx=aa2+x2$

Differentiation exercise 10.3 question 7

Answer:
dydx=21−x2
Hint:
ddx(Constant)=0;∂dx(xn)=nxn−1
Given:
sin−1⁡(2x2−1),0<x<1
Solution:
$\begin {array}{l} Let,\ y=\sin ^{-1}\left(2 x^{2}-1\right)\\ Let \ \mathrm{x}=\cos \theta\\ Now,\ y=\sin ^{-1}\left\{2 \cos ^{2} \theta-1\right\}\\ Using\\ 2 \cos ^{2} \theta-1=\cos 2 \theta\\ y=\sin ^{-1}(\cos 2 \theta)\\ y=\sin ^{-1}\left\{\sin \left(\frac{n}{2}-2 \theta\right)\right\} \end{}$
As we know,
sin⁡(n2−θ)=cos⁡θ
Considering the limits, 0<x<1
$\begin {array}{l} 0<\cos \theta<1\\ 0<\theta<\frac{n}{2} \ \ \ \ \ \ \left\{\cos \frac{\pi}{2}=0, \cos 0=1\right\}\\ 0>-2 \theta>-\pi\\ \frac{n}{2}>\frac{n}{2}-2 \theta>-\frac{\pi}{2}\\ \text { Now, } \mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}-20}{2}\right)\right\}\\ \mathrm{y}=\frac{\mathrm{n}}{2}-2 \theta \ \ \ \ \ \ \ \ \ \left\{\sin ^{-1}(\sin \theta)=\theta\right\} \text { if } \theta \in\left[-\frac{n}{2}, \frac{n}{2}\right]\\ \therefore \mathrm{x}=\cos \theta, \theta=\cos ^{-1} \mathrm{x}\\ y=\frac{n}{2}-2 \cos ^{-1} x \end{}$
Differentiating with respect ts? x, weget
dydx=d∂x(n2−2cos−1⁡x)
As we know
$\begin {array}{l} \frac{d}{d x}(constant)=0, \frac{\partial}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=0-2\left(-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}} \end{}$

Differentiation exercise 10.3 question 8

Answer :dydx=−21−x2

Hint:
∂dx(Constant)=0;∂dx(xn)=nxn−1

Given:
sin−1⁡(1−2x2),0<x<1
Solution:
$\begin {array} {ll}Let \ y=\sin ^{-1}\left\{1-2 x^{2}\right\}\\ Let \ x=\sin \theta\\ Then,\ 0=\sin ^{-1} x\\ y=\sin ^{-1}\left\{-2 \sin ^{2} \theta\right\}\\ Using,\ 1=2 \sin ^{2} \theta=\cos 2 \theta\\ \mathrm{y}=\sin ^{-1}\{\cos 2 \theta\}\\ \mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}}{2}-2 \theta\right)\right\} \end{}$
As we know,
Unknown environment ''
Considering the limits,
$\begin {array} {ll}0<x<1\\ 0<\sin \theta<1\\ 0<\theta<\frac{n}{2}\ \ \ \ \ \ \ \left\{\sin \frac{1}{2}=1, \sin \theta=0\right\}\\ 0 \angle 2 \theta=\pi\\ 0>-2 \theta>-\pi\\ \frac{1}{2}>\frac{\pi}{2}-2 \theta>-\frac{n}{2}\\ \end{}$
Now,
$\begin {array} {ll}\mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}}{2}-2 \theta\right)\right\} \\ y=\frac{n}{2}-2 \theta \\ y=\frac{n}{2}=2 \sin ^{-1} x \end{}$
Differentiating with respect to x,
$dydx=ddx(n2−2sin−1⁡x) As we know ddx( constant )=0ddx(sin−1⁡x)=11−x2dydx=0−211−x2dydx=−21−x2$

Differentiation exercise 10.3 question 9

Answer :

dydx=−aa2+x2
Hint:
ddx(Constant)=0ddx(xn)=nxn−1
Given:
cos−1⁡{xx2+a2}
Solution:
$Let,\\$
$\mathrm{y}=\cos ^{-1}\left\{\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)\\$
$Let\\$
$\mathrm{x}=\operatorname{acot} \theta\\$
$\theta=\cot ^{-1}\left(\frac{x}{a}\right)\\$
$ Now y=cos−1⁡{acot⁡θa2cot2⁡θ+a2}y=cos−1⁡{acot⁡θa2(cot2⁡θ+1)}y=cos−1⁡{acot⁡θacot2⁡θ+1} Using 1+cot2⁡θ=cosec2⁡θy=cos−1⁡{acot⁡θacosec2⁡θ}y=cos−1⁡{acot⁡θacosecθ}y=cos−1⁡{cot⁡θcosec⁡θ}$
As we Know,
$cot⁡θ=cos⁡θsin⁡θ , cosec⁡θ=1sin⁡θy=cos−1⁡{cos⁡θsin⁡θ×sin⁡θ}y=cos−1⁡(cos⁡θ)y=θy=cot−1⁡(xa) Differentiating with respect to x We get dydx=ddx(cot−1⁡(xa))∴∂dx(cot−1⁡x)=−11+x2dydx=−11+(xa)2×1a$
$dydx=−11+x2a2×1adydx=−1a2+x2a2×1adydx=a2a2+x2×1adydx=−aa2+x2$

Differentiation Exercise 10.3 Question 10

Answer: dydx=1
Hint:
ddx (constants) =0
ddx(xn)=nxn−1
Given:
sin−1⁡{sin⁡x+cos⁡x2}
−3π4<x<π4
Soluton:
Let,y=sin−1⁡{sin⁡x+cos⁡x2}
Now,
y=sin−1⁡{sin⁡x12+cos⁡x12}
y=sin−1⁡{sin⁡xcos⁡(π4)+cos⁡xsin(π4)}
sin⁡(π4)=12&
cos⁡(π4)=12
Using,
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B
y=sin−1⁡{sin⁡(x+π4)}
Considering the limits,
−3π4<x<π4
−3π4+π4<x+π4<π4+14
−π2<x+π4<π2
Now,
y=sin−1⁡{sin⁡(x+π4)}
y=x+π4 {sin−1⁡(sin⁡θ)=θ, if θε[π2,π2]}
Differentiating with respect to x , We get
dydx=1+0
ddx (constant) =0
dydx=1

Differentiation Exercise 10.3 Question 11

Answer: dy dx=−1
Hint:
ddx (constants) =0
ddx(xn)=nxn−1
Given:
cos−1⁡{cos⁡x+sin⁡x2}
−π4<x<π4
Solution:
Let,
y=cos−1⁡{cos⁡x+sin⁡x2}
Now
y=cos−1⁡{cos⁡x12+sin⁡x12}
y=cos−1⁡{cos⁡xcos⁡(π4)+sin⁡xsin⁡(π4)}
sin⁡(π4)=12&
cos⁡(π4)=12
Using,
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B
y=cos−1⁡{cos⁡(x−π4)}
Considering the limits,

−π4<x<π4
−π4−π4<x−π4<π4−π4
−2π4<x−π4<0
−π2<x−π4<0
Now,
y=cos−1⁡{cos⁡(x−π4)}
y=−(x−π4) $\left\{cos−1⁡(cos⁡θ)=−θ if θ∈[−π,0]\right\}$
Differentiating with respect to x , We get

dydx=−1

ddx( constants )=0

Differentiation Exercise 10.3 Question 12 (i)

Answer:
dydx=121−x2
Hint:
ddx(constant)=0
ddx(xn)=nxn−1
Given:
tan−1⁡{x1+1−x2}
−1<x<1
Solution:
Let,
$y=tan−1⁡{x1+1−x2}x=sin⁡θθ=sin−1⁡x$
Now,y=tan−1⁡{sin⁡θ1+1sin2⁡θ}
Using sin2⁡θ+cos2⁡θ=1
$y=tan−1⁡{sin⁡θ1+cos2⁡θ}y=tran−1⁡{sin⁡θ1+cos⁡θ} Using 2cos2⁡θ=1+cos⁡2θ2sin⁡θcos⁡θ=sin⁡2θ$$y=tan−1⁡{2sin⁡θ2cos⁡θ22cos2⁡θ2}y=tan−1⁡{sin⁡θ2cos⁡θ2}tan⁡θ=sin⁡θcos⁡θy=tan−1⁡{tan⁡θ2}$

Considering the limits,

$−1<x<1−1<sin⁡θ<1−π2<θ2<π2−π4<θ2<π4$

Now,

$y=tan−1⁡{tan⁡θ2}y=θ2$ tan−1⁡(tan⁡θ)=0 if θ∈[−π2,π2]

y=12sin−1⁡x

Differentiating with respect to x , We get

$dydx=−1ddx( constants )=0dydx=12(12ddx(sin−1⁡x)ddx(sin−1⁡x)=11−x2dydx=1211−x2dydx=121−x2$


−1<x<1

Differentiation Exercise 10.3 Question 12 (ii)

Answer:
dydx=12
Hint:
cos⁡2x=2cos2⁡x−1
tan⁡(π2−θ)=cot⁡θ
Given:
tan−1⁡{1+cos⁡xsinx}
Solution:
$y=tan−1⁡{1+cos⁡xsin⁡x}y=tan−1⁡{2cos2⁡x22sin⁡x2cos⁡x2}y=tan−1⁡(cot⁡x2)y=tan−1⁡(tan⁡(π2−x2))$ [(π2−θ)=cot⁡θ]
y=x2
Differentiating w.r.t x
dydx=12

Differentiation Exercise 10.3 Question 13

Answer:dydx=12a2−x2
Hint:
$ddx( constants )=0ddx(xn)=nxn−1$
Given:
$tan−1⁡{xa+a2−x2}−a<x<−a$
Solution:
Let,
x=asin⁡θ
θ=sin−1⁡xa
Now,
$y=tan−1⁡{asin⁡θa+a2−a2sin2⁡θ}y=tan−1⁡{asin⁡θa+a1−sin2⁡θ}$
Using
$cos2⁡θ+sin2⁡θ=1y=tan−1⁡{asin⁡θa+acos⁡θ}y=tan−1⁡{asin⁡θa+acos⁡θ}y=tan−1⁡{asin⁡θa(1+los⁡θ)}Using⁡2cos2⁡θ=1+cos⁡θy=tan−1⁡{sin⁡θ1+cos⁡θ} and 2sin⁡θcos⁡θ=sin⁡2θy=tan−1⁡{sin⁡θ1+cos⁡θ}$
$y=tan−1⁡{2sin⁡θ2cos⁡θ22cos2⁡θ2}y=tan−1⁡{sin⁡θ2cos⁡θ2}y=tan−1⁡(tan⁡θ2)$ ∴sin⁡θcos⁡θ=tan⁡θ
Considering the limit

$−a<x<a−1<sin⁡θ<1−π2<θ<π2−π4<θ2<π4$ {sin⁡π2=1}
Now,$y==tan−1⁡{tan⁡θ2}y=θ2y=12sin−1⁡xa$

Differentiating with respect to x , We get

$dydx=ddx(12sin−1⁡xa)∴ddx(sin−1⁡x)=11−x2dydx=1211−(x)2×1a⇒12×11−x2a2×1a⇒dydx=12×1a2−x2a2×1adydx=12a2−x2$

Differentiation Exercise 10.3 Question 14


Answer:dydx=11−x2
Hint:
$ddx( constants )=0ddx(xn)=nxn−1$
Given:
$sin−1⁡{x+1−x22}−1<x<1$
Solution:
Let
y=sin−1⁡{x+1−x22}
Let
x=sin⁡θ
θ=sin−1⁡x
Now,
y=sin−1⁡{sin⁡θ+1−sin2⁡θ2}
Using
$sin2⁡θ+cos2⁡θ=1y=sin−1⁡{sin⁡θ+cos2⁡θ2]y=sin−1⁡{sin⁡θ+cos⁡θ2}$
Now,
$y=sin−1⁡{sin⁡θ12+cos⁡θ12}y=sin−1⁡{sin⁡θcos⁡(π4)+cos⁡θsin⁡(π4)}sin⁡(π4)=12&cos⁡(π4)$
y=sin−1⁡{sin⁡(θ+π4)}

Considering the limits,

$−1<x<1−1<sin⁡θ<1−π2<θ<π2$ {sin⁡π2=1}

$−π2+π4<θ+π4<π2+π4−π4<θ+π4<3π4$

Now,

$y=sin−1⁡{sin⁡(θ+π4)}y=θ+π4y=sin−1⁡x+π4$ sin−(sin⁡θ)=θifθε[−π2,π2]Differentiating with respect to x , We get
$ddx(sin−1⁡x)=11−x2ddx (constant) =0dydx=11−x2$

Differentiation Exercise 10.3 Question 15

Answer:dydx=−11−x2
Hint:
$ddx( constants )=0ddx(xn)=nxn−1$
Given:
$cos−1⁡{x+1−x22}−1<x<1$
Solution:
$y=\cos ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\} \$
Let,
x=sin⁡θ
θ=sin−1⁡x
$y=cos−1⁡{sin⁡θ+1−sin2⁡θ2}y=cos−1⁡{sin⁡θ+cos2⁡θ2} Using sin2⁡θ+cos2⁡θ=1y=cos−1⁡{sin⁡θ+cos⁡θ2}y=cos−1⁡{sin⁡θ12+cos⁡θ12}y=cos−1⁡{sin⁡θsin⁡(π4)++cos⁡θcos⁡(π4)}$
∴sin⁡(π4)=12
cos⁡(π4)=12
Using,
$cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡By=cos−1⁡{cos⁡(θ−π4)}$
Considering the limits,
$−1<x<1−1<sin⁡θ<1−π2<θ<π2−π2−π4<θ−π4<12−π4−3π4<8−π4<π4$
Now,
$y=cos−1⁡{cos⁡(θ−π4)}y=−(θ−π4)$ {∴cos−1⁡(cos⁡θ)=−0 if θε[−π,0]}
$y=−(θ−π4)y=−sin−1⁡x+π4$ [sincex=sin⁡θ]
Differentiating with respect to x , We get
$dydx=ddx(−sin−1⁡x+π4)∴ddx( constants )=0ddx(sin−1⁡x)=11−x2dydx=−11−x2+0dydx=−11−x2$

Differentiation Exercise 10.3 Question 16

Answer: dydx=41+4x2
Hint:
$ddx( constants )=0ddx(xn)=nxn−1$
Given:
$tan−1⁡{4x1−4x2}−12<x<12$
Solution:
Let,
y=tan−1⁡{4x1−4x2}
Let,
2x=tan⁡θ
θ=tan−1⁡2x
y=tan−1⁡{2tan⁡θ1−tan2⁡θ}
Using tan⁡2θ=2tan⁡θ1−tan2⁡θy=tan−1⁡(tan⁡2θ)

Considering the limits,

$−12<x<12−1<2x<1−1<tan⁡θ<1−π4<θ<π4−π2<2θ<π2$ {tan⁡45∘=1}

Now,

y=tan−1⁡(tan⁡2θ)

y=2θ

y=2tan−1⁡(2x) [since,2x=tan⁡θ]

Differentiating with respect to x , We get

$dydx=ddx(2tan−1⁡2x)dydx=2×21+(2x)2dydx=41+4x2$ d(tan−1x)dx=11+x2

Diffrentiation Exercise 10.3 Question 17

Answer: dydx=2x+1log21+4x
Hint:
$ddx( constants )=0ddx(xn)=nxn−1$
Given:
$tan−1⁡[2x+11−4x]−∞<x<0$
Solution:
y=tan−1⁡[2x+11−4x]
Let,
2x=tan⁡θ
y=tan−1⁡{2x2x1−(2x)2}
As we Know
$am+n→am⋅any=tan−1⁡{2tan⁡θ1−tan2⁡θ}$
Using
tan⁡2θ=2tan⁡θ1−tan2⁡θy=tan−1⁡(tan⁡2θ)

Considering the limits,

$−∞<x<02−∞<2x<200<tan⁡θ<10<π4$ {20=1,2−∞=0}

0<2θ<π2 {tan⁡π4=1}

0<2θ<π2

Now

y=tan−1⁡(tan⁡θ)

y==2θ

y=2tan−1⁡(2x) $\left\{sin⁡ce2x=tan⁡θθ=tan−1⁡(2x)\right\}$

Differentiating with respect to x , We get

$dydx=ddx(2tan−1⁡(2x))dydx=2×2xlog⁡21+(2x)2dydx=2x+1log⁡21+4x$ $∴d(tan−1⁡x)dx=11+x2ddx(2x)=log⁡2{am⋅an⇒am+h}$

Differentiation Exercise 10.3 Question 18

Answer:dydx=2axloga1+a2x
Hint:
dcdx=0;ddx(xn)=nxn−1
Given:
tan−1⁡{2ax1−a2x},a>1,−∞<x<0
Solution:
Let
y=tan−1⁡{2ax1−a2x},
Let,
$Missing \end{aligned}\ &\text { Considering the limits, }\ &-\infty<\mathrm{x}<0\ &a^{-\infty}<a^{x}<a^{0} \end{aligned}$
0<tan⁡θ<1 {ao=1}
0<θ<π4 {tan⁡(π4)=1}
Now,y=tan−1⁡(tan⁡2θ)
y=2θ
y=2tan−1⁡(ax) sinceax=tan⁡θ,θ=tan−1⁡(ax)
Differentiating with respect to x, we get
$dydx=ddx(2tan−1⁡(ax))dydx=2×ax1+(ax)2log⁡addx(tan−1⁡x)=11+x2dydx=2axlog⁡a1+a2x$

Differentiation Exercise 10.3 Question 19

Answer:dydx=−121−x2
Hint:
$ddx( constants )=0ddx(xn)=nxn−1$
Given:
sin−1⁡{1+x+1−x2},0<x<1
Solution:
Let
y=sin−1⁡{1+x+1−x2}
Let,x=cos⁡2θ
$ Now, sin−1⁡{1+cos⁡2θ+1−cos⁡2θ2} Using 1−2sin2⁡θ=cos⁡2θ2cos2⁡θ−1=cos⁡2θy=sin−1⁡{2cos2⁡θ+2sin2⁡θ2}$
Now,
$y=sin−1⁡{2cos⁡θ+2sin⁡θ2}y=sin−1⁡{cos⁡θ12+sin⁡θ12}y=sin−1⁡{sin⁡θcos⁡(π4)+cos⁡θsin⁡(π4)}∴sin⁡(π4)=12&cos⁡(π4)=12Usin⁡sin(A+B)=sin⁡Acos⁡B+cos⁡AsinBy=sin−1⁡{sin⁡(θ+π4)}$
Considering the limits,
$0<x<10<cos⁡2θ<10<2θ<π20<θ<π40+π4<(θ+14)<π2+π4π4<(θ+π4)<π2$
Now,y=sin−1 {sin⁡(θ+π4)}
y=θ+π4 sin−1⁡(sin⁡θ)=θ,ifθε[−π2,π2]
y=12cos−1⁡x+π4 {sincex=cos⁡2θ}
dydx=ddx(12cos−1⁡x+π4)
$ddx( constant )=0 ddx(cos−1⁡x)=−11−x2dydx=12(−11−x2)+0dydx⇒−121−x2$

Differentiation Exercise 10.3 Question 20

Answer: dydx=a2(1+a2x2)
Hint:
$ddx( constant )=0ddx(xn)=nxn−1$
Given:
$tan−1⁡{1+a2x2−1ax},x≠0$
Solution:
Let,
y=tan−1⁡{1+a2x2−1ax}
Letax=tan⁡θ
Now
y=tan−1⁡{1+tan2⁡θ−1tan⁡θ}
Using
$sec2⁡θ=1+tan2⁡θy=tan−1⁡{sec2⁡θ−1tan⁡θ}y=tan−1⁡{sec⁡θ−1tan⁡θ}y=tan−1⁡{1cos⁡θ−1sin⁡θcos⁡θ}$
$ Using tan⁡θ=sin⁡θcos⁡θ,sec⁡θ=1cos⁡θy=tan−1⁡{1−cos⁡θcos⁡θsin⁡dcos⁡θ}y=tan−1⁡{1−cos⁡θcos⁡θ×cos⁡θsin⁡θ}y=tan−1⁡{1−cos⁡θsin⁡θ}$
$ Using 2sin2⁡θ=1−cos⁡2θ and 2sin⁡θcos⁡θ=sin⁡2θy=tan−1⁡{2sin2⁡θ/22sin⁡θ/2cos⁡θ/2}y=tan−1⁡{tan⁡θ2}$
y=θ2
y=12tan−1ax
Differentiating with respect to x , We get
$dydx=ddx(12tan−1⁡x) Using ddx(tan−1⁡x)⇒11+x2dydx=12×11+(ax)2×1adydx=a2(1+a2x2)$

Differentiation Exercise 10.3 Question 21

Answer:dydx=12
Hint:
dydx(xn)=nxn−1
dydx(constant)=0
Given:
tan−1⁡{sin⁡x1+cos⁡x},−π<x<π
Solution:
Let,
y=tan−1⁡{sin⁡x1+cos⁡x}
Function y is defined for all the real numbers where cos⁡x≠−1
$ Using 2cos2⁡θ=1+cos⁡2θ2sin⁡θcos⁡θ=sin⁡2θy=tan−1⁡{2sin⁡x2cos⁡x22cos2⁡x2}y=tan−1⁡{−sin⁡x/2cos⁡x/2}y=tan−1⁡{tan⁡x2}y=x2$
Differentiating with respect to x , We get
dydx=ddx(x2)
dydx=12

Differentiation Exercise 10.3 Question 22

Answer: dydx=−11+x2
Hint:
$ddx( Constant )=0ddx(xn)=nxn−1$
Given:
sin−1⁡{11+x2}
Solution:
Let,
y=sin−1⁡{11+x2}
Let
x=cot⁡θ
θ=cot−1⁡x
Now,
y=sin−1⁡{11+cot2⁡θ}
Using,
$1+cot2⁡θ=cosec2⁡θy=sin−1⁡{1cosec2⁡θ}y=sin−1⁡{1cosec⁡θ}$ {1cosecθ=sin⁡θ}
y=sin−1⁡(sin⁡θ) sin−1⁡(sin⁡θ)=sin⁡θ,if θε[−π2,π2]
y=θy=cot−1⁡x

Differentiating with respect to x , We get

$dydx=ddx(cot−1⁡x)∴ddx(cot−1⁡x)⇒−11+x2dydx=−11+x2$

Differentiation Exercise 10.3 Question 23

Answer:
dydx=2nxn−11+x2n
Hint:
$ddx( constsant )=0ddx(xn)=nxn−1$
Given:
$cos−1⁡{1−x2n1+x2n}0<x<∞$
Solution:
y=cos−1{1−x2n1+x2n}
Let,
xn=tan⁡θ
θ=tan−1⁡(xn)
y=cos−1⁡{1−tan2⁡θ1+tan2⁡θ}y=cos−1⁡{cos⁡2θ}

Using,

1−tan2⁡θ1+tan2⁡θ=cos⁡2θ

Considering the limits,

$0<x<∞0<xn<∞0<θ<π2$

Now, {cos−1⁡(cos⁡θ)=θ,ifθε[0,π]}

y=cos−1⁡(cos⁡2θ)

y=2θ Sincexn=tan⁡θ
Differentiating with respect to x we get
dydx=ddx(2tan−1⁡(xn))
Using
$ddx(tan−1⁡x)=11+x2dydx=2×11+(xn)2×nxn−1dydx=2nxn−11+x2n$

Differentiation Exercise 10.3 Question 24

Answer:dydx=0
Hint:
$ddx( constsant )=0ddx(xn)=nxn−1$
Given:
sin−1⁡(1−x21+x2)+sec−1⁡(1+x21−x2)
Solution:
Let,
y=sin−1⁡(1−x21+x2)+sec−1⁡(1+x21−x2)
Using
$sec−1⁡x=cos−1⁡(1x)y=sin−1⁡(1−x21+x2)+cos−1⁡(1−x21+x2)$
Using$sin−1⁡x+cos−1⁡x=π2y=π2$

Differentiating with respect to x we get

dydx=ddx(π2)

dydx=0 {ddx(constant)=0}

Differentiation Exercise 10.3 Question 26

Answer:12x(1+x)
Hint:
ddx(constant)=0;
ddx(xn)=nxn−1
Given:
tan−1⁡(x+a1−xa)
Solution:
Let,
y=tan−1⁡(x+a1−xa)
Since,tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)
y=tan−1⁡x+tan−1⁡a
Differentiating it with respect to x using chain qule.
dydx=ddx(tan−1⁡x)+ddx(tan−1⁡a)
Since ,
$ddx(tan−1⁡x)=11+x2;ddx(xn)=nxn−1dydx=11+(x)2ddx(x)+0dydx=11+(x1/2)2×12(x)12−1dydx=11+x×12(x)−12dydx=11+x×12xdydx=12x(1+x)$

Differentiation Exercise 10.3 Question 27

Answer:dydx=1
Hint:
ddx(constant)=0
ddx(xn)=nxn−1
Given:
tan−1⁡[a+btanxb−atanx]
Solution:
Let,
$y=tan−1⁡[a+btan⁡xb−atan⁡x]y=tan−1⁡[a+btan⁡xbb−atanxb]y=tan−1⁡[ab+tan⁡x1−abtan⁡x]$
$y=tan−1⁡[tan⁡(tan−1⁡ab)+tan⁡x1−tan⁡(tan−1⋅gb)tan⁡x]y=tan−1⁡[tan⁡(tan−1⁡ab+x)]∴tan⁡(A+B)=tan⁡A+tan⁡B→1−tan⁡Atan⁡By=tan−1⁡(ab)+x$
Differentiating it with respect to x , We get
$dydx=ddx(tan−1⁡(ab))+ddx(x)ddx( constant )=0ddx(tan−1⁡x)=11+x2dydx=11+(ab)2×0+1dydx=0+1dydx=1$

Differentiation Exercise 10.3 Question 28

Answer:dydx=11+x2
Hint:
$ddx( constant )=0;ddx(xn)=nxn−1$
Given:
tan−1⁡(a+bxb−ax)
Solution:
Let,
$y=tan−1⁡(a+bxb−ax)y=tan−1⁡[a+bxbb−axb]$

Since,tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)

Differentiating it with respect to x , We get

$dydx=ddx(tan−1⁡(ab)+tan−1⁡x Since, ddx(tan−1⁡x)=11+x2dydx=11+(ab)2×0+11+x2dydx=0+11+x2dydx=11+x2$

Differentiation Exercise 10.3 Question 29

Answer:dydx=aa2+x2
Hint:
$ddx( constan t)=0;ddx(xn)=nxn−1$
Given:
tan−1⁡(x−ax+a)
Solution:
Let,
$y=tan−1⁡(x−ax+a)y=tan−1⁡(x−axx+ax)$$y=tan−1⁡(xx−axxx+ax)y=tan−1⁡(1−ax1+1×ax)y=tan−1⁡(1)+tan−1⁡(ax)$

Using,

tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)

Differentiating its with respect to xusing chain tule.

$dydx=0−11+(ax)2ddx(ax)dydx=−11+a2x2(−ax2)dydx=−1x2+a2x2×(−ax2)dydx=−x2x2+a2×(−ax2)$ Using ddx(xn)=nxn−1

ddx=ax2+a2 ddx(constant)=0

Differentiation Exercise 10.3 Question 30

Answer: dydx=31+9x2−21+4x2
Hint:
$ddx( constan t)=0;ddx(xn)=nxn−1$
Given:
tan−1⁡(x1+6x2)
Solution:
Let, y=tan−1⁡(x1+6x2)
y=tan−1⁡(3x−2x1+(3x)(2x))
Since,3x−2x=x
$y=tan−1⁡3x−tan−1⁡2x[ Since +tan⁡x−tan−1⁡y=tan−1⁡(x−y1+xy)]$
Differentiating it with respect tox using chain rule,
$dydx=11+(3x2)ddx(3x)−11+(2x)2ddx(2x){ Since ⇒ddxtan−1⁡(x)=11+x2}dydx=11+9x2(3)−11+4x2(2)dydx=31+9x2−21+4x2$

Differentiation Exercise 10.3 Question 31


Answer:dydx=31+9x2+21+4x2
Hint:
ddx(constant)=0;
ddx(xn)=nxn−1
Given:
tan−1⁡(5x1−6x2)
Solution:
Let,
$y=tan−1⁡(5x1−6x2)y=tan−1⁡(3x+2x1−(3x)(2x))$
$ By using 5x=3x+2xy=tan−1⁡(3x)+tan−1⁡(2x)$
 Since, tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)
Differentiating it with respect to x using chain rule,
dydx=ddx(tan−1⁡(3x))+ddx(tan−1⁡(2x))
Using
$ddx(tan−1⁡x)=11+x2dydx=11+(3x)2ddx(3x)+11+(2x)2ddx(2x)dydx=11+9x2(3)+11+4x2(2)dydx=31+9x2+21+4x2$

Differentiation Exercise 10.3 Question 32

Answer: dydx=1
Hint:
$ddx( constan t)=0;ddx(xn)=nxn−1$
Given:
tan−1⁡[cos⁡x+sin⁡xcos⁡x−sin⁡x]
Solution:
$y=tan−1⁡[cos⁡x+sin⁡xcos⁡x−sin⁡x]y=tan−1⁡[cos⁡x+sin⁡xcos⁡xcos⁡x−sin⁡xcos⁡x]y=tan−1⁡[1+tan⁡x1−tan⁡x] since, sin⁡xcos⁡x=tian⁡x$
$y=tan−1⁡[tan⁡π4+tan⁡x1−tan⁡π4tan⁡x]y=tan−1⁡[tan⁡(π4+x)] Since tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡tan⁡By=π4+x$
Differentiating it with respect to x,
dydx=0+1 $\left\{ddx( constant )=0ddx(x)=1\right\}$
dydx=1

Differentiation Exercise 10.3 Question 33

Answer: dydx=13x2(1+x23)
Hint:
$ddx( constant )=0;ddx(xn)=nxn−1$
Given:
tan−1⁡[x1/3+a1/31−(ax)1/3]
Solution:
Let,
$y=tan−1⁡[x13+a1/31−(ax)1/3]y=tan−1⁡(x1/3)+tan−1⁡(a1/3) Since, tan−1⁡x+tan−1⁡y=tan−1⁡(x+y1−xy)$
Differentiating it with respect to x using chain rule,
$dydx=11+(x1/3)2ddx(x1/3)+11+(a1/3)2ddx(a1/3)dydx=11+x2/3×(13x13−1)+11+(a1/3)2×0dydx=11+x2/3×13x−2/3dydx=13x2/3(1+x2/3)$
Using
$ddx(tan−1⁡x)=11+x2ddx(xn)=nxn−1ddx( constant )=0$

Differentiation Exercise 10.3 Question 34

Answer:dydx=2x+1log21+4x
Hint:
ddx(constant)=0
∂∂x(xn)=nxn−1
Given:
sin−1⁡(2x+11+4x)
Solution:
Let
y=sin−1⁡(2x+11+4x)
To find the domain we need to find all x such that
−1≤2x+11+4x≤1
Since the quantity in the middle is always positive, we need to find
all such that 2x+11+4x≤1
ie, all x such that 2x+1=1+4x
2≤12x+2x, which is true for a x
Hence the function is defined at all real numbers
Putting 2x=tan⁡θ
$y=sin−1⁡(2x+11+4x)y=sin−1⁡(2x⋅21+(2x)2)y=sin−1⁡(2tan⁡θ1+tan2⁡θ) Using sec2⁡θ=1+tan2⁡θ$
$y=sin−1⁡(2tan⁡θsec2⁡θ)y=sin−1⁡(2sin⁡θcos⁡θ×cos2⁡θ)\;Since,tan⁡θ=sin⁡θcos⁡θ′sec⁡θ,1sec=cos⁡θy=sin−1⁡(2sin⁡θcos⁡θ)Usin⁡sin=sin⁡2θ=2sin⁡θcos⁡θ$
y=sin−1⁡(sin⁡θ) $\left\{sin−1⁡(sin⁡θ)=θ if θ∈[−12,02]\right\}$
y=2θ
y=2tan−1⁡(2x) $\left\{ since 2x=tan⁡θ\boldsymbolθ=tan−1⁡(2x)\right\}$
$dydx⇒ddx(2tan−1⁡(2x))dydx=2×11+(2x)2∂dx(2x)dydx=2×11+4x⋅(2x)log⁡2dydx⇒2x+1log⁡21+4x$  Since ddx(tan−1⁡x)=11+x2

Differentiation Exercise 10.3 Question 35

Answer: Hence Provedydx=41+x2
Hint:
$ddx( constan t)=0;ddx(xn)=nxn−1$
Given:
$y=sin−1⁡(2x1+x2)+sec−1⁡(1+x21−x2)0<x<1$
Solution:
Prove :dydx=41+x2
$y=sin−1⁡(2x1+x2)+cos−1⁡(1−x21+x2) Since sec−1⁡x=cos−1⁡(1x)y=sin−1⁡(2x1+x2)+cos−1⁡(1−x21+x2)$
Let,
x=tan⁡θ
θ=tan−1⁡x
$y=sin−1⁡(2tan⁡θ1+tan2⁡θ)+cos−1⁡(1−tan2⁡θ1+tan2⁡θ)y=sin−1⁡(2tan⁡θsec2⁡θ)+cos−1⁡(cos⁡2θ)$
Using
$sec2⁡θ=1+tan2⁡θ1sec⁡θ=cos⁡θ1−tan2⁡θ1+tan2⁡θ=cos⁡2θ$
$y=sin−1⁡(2sin⁡θcos⁡θ×cos2⁡θ)+cos−1⁡(Cos⁡2θ)y=sin−1⁡(2sin⁡θcos⁡θ)+cos−1⁡(cos⁡20)y=sin−1⁡(sin⁡2θ)+cos−1⁡(cos⁡2θ)−(i)using⁡2sin⁡θcos⁡θ=sin⁡2θ$
Considering Limits,
$0<x<10<tan⁡θ<10<θ<π40<(2θ)<π2$
so from ei(i)
y=2θ+2θ
y=4θ
$cos−1⁡(cos⁡θ)⇒ if θ∈[0,π]y=4tan−1⁡x$ since,x=tan⁡θ,θ=tan−1⁡x
Differentiating it with respect to x
$dydx=4ddx(tan−1⁡x) Using ddx(tan−1⁡x)=11+x2dydx=4(−11+x2)dydx=41+x2$

Differentiation Exercise 10.3 Question 36

Answer: Hence Prove ,dydx=21+x2
Hint:
$ddx( constan t)=0;ddx(xn)=nxn−1$
Given:
$y=sin−1⁡(x1+x2)+cos−1⁡(11+x2)0<x<∞$
Solution:
Let,
x=tan⁡θ,
θ=tan−1⁡x
$y=sin−1⁡(tan⁡θ1+tan2⁡θ)+cos−1⁡(11+tan2⁡θ)y=sin−1⁡(tan⁡θsec2⁡θ)+cos−1⁡(1sec2⁡θ) Using sec2⁡θ=1+tan2⁡θsin⁡θcos⁡θ=tan⁡θ1sec⁡θ=θcos⁡θ$
$y=sin−1⁡(tan⁡θsec⁡θ)+cos−1⁡(1sec⁡θ)y=sin−1⁡(sin⁡θcos⁡θ×cos⁡θ)+cos−1⁡(cos⁡θ)y=sin−1⁡(sin⁡θ)+cos−1⁡(cos⁡θ)−(i)$
Considering limit

$0<x<∞0<tan⁡θ<∞0<θ<π2$
So From eq (i)
y=θ+θ  Since, sin−1⁡(sin⁡θ)=θ if θε[−π2,π2]
y=2θ cos−1⁡(cos⁡θ)=θ if θε[0,π]
y=2tan−1⁡x [sincex=tan⁡θ]

Differentiating it with respect to x

$ddx(tan−1⁡x)=11+x2dydx=2×11+x2dydx=21+x2$

Differentiation Exercise 10.3 Question 37(i)

Answer: dydx=−1
Hint:
$ddx( constan t)=0ddx(xn)=nxn−1$
Given:
cos−1⁡(sin⁡x)
Solution:
Let,
y=cos−1⁡(sin⁡x)
We observe that this function is defined for all real numbers
$y=cos−1⁡(sin⁡x)y=cos−1⁡[cos⁡(π2−x)]y=π2−x$ $\left\{since⁡cos−1⁡(cos⁡θ)=θ if θ∈[θ,π]\right\}$
Differentiating it with respect to x,
$dydx=0−1dydx=−1$ {Since,d(constant)dx=0}

Differentiation Exercise 10.3 Question 37 (ii)

Answer: dydx=11+x2
Hint:
$ddx( constan t)=0ddx(xn)=nxn−1$
Given:
cos−1⁡(1−x1+x)
Solution:
y=cos−1⁡(1−x1+x)
Let,
$x=tan⁡θθ=tan−1⁡xy=cot−1⁡(1−tan⁡θ1+tan⁡θ)y=cot−1⁡(tan⁡n4−tan⁡θ1+tan⁡n4tan⁡θ)$
$y=cot−1⁡[tan⁡(n4−θ)] Using, tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡By=cot−1⁡[tan⁡(π4−θ)]y=cot−1⁡[cots⁡(π2−(π−θ4−θ)]$ ∴sincecot⁡(π2−θ)=tan⁡θ
$y=cot−1⁡[cot⁡(π2−π4+θ)]y=cot−1⁡[cot⁡(π4+θ)]y=π4+θ$ {cot−1⁡(cot⁡θ)=θ, if θ<[−π2,π2]}
y=π4+tan−1⁡x $\left\{ Since x=tan⁡θθ=tan−1⁡x\right\}$
Differentiating it with respect to x,
$dydx=0+11+x2ddx(tan−1⁡x)=11+x2∂dx( constant )=0dydx=11+x2$

Differentiation Exercise 10.3 Question 38

Answer:dydx=12, Hence,dydx is independent of x
Hint:
$ddx (constant) =0;ddx(xn)=nxn−1$
Given:
y=cot−1⁡{1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x}
Solution:
$y=cot−1⁡{1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x} Then, 1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x×1+sin⁡x+1−sin⁡x1+sin⁡x+1−sin⁡x$
$(1+sin⁡x+1+sin⁡x)2(1+sin⁡x)−(1−sin⁡x) Using (a−b)(a+b)=a2−b2(a+b)2=a2+b2+2ab(1+sin⁡x)+(00(1−sin⁡x)+2(1−sin⁡x)(1+sin⁡x)(1+sin⁡x)−(1−sin⁡x)1+sin⁡x+1−sin⁡x+2(1−sin⁡x)(1+sin⁡x)1+sin⁡x−1+sin⁡x$
$⇒2+21+sin⁡x−sin⁡x−sin2⁡x2sin⁡x⇒2+21−sin2⁡x2sin⁡x⇒2(1+1−sin2⁡x)2sin⁡x⇒1+1sin2⁡xsin⁡x$
Using cos2⁡x+sin2⁡x=1
$⇒1+cos2⁡xsin⁡x⇒1+cos⁡xsin⁡x Using sin⁡2θ=2sin⁡θcos⁡θ1+Cos⁡2θ=2cos2⁡θ⇒2cos2⁡x22sin⁡x2cos⁡x2⇒cot⁡x2$ {cos⁡xsin⁡x=cot⁡x}

Therefore, equation (1) becomes

$y=cot−1⁡(cot⁡x2)y=x2$ {cot−1⁡(cot⁡θ)=θifθε[−π2,π2]}

Differentiating it with respect to x

dydx=12ddx(x)

dydx=12

Differentiation Exercise 10.3 Question 39

Answer: dydx=41+x2
Hint:
$ddx( constiant )=0ddx(xn)=nxn−1$
Given:
$y=tan−1⁡(2x1−x2)+sec−1⁡(1+x21−x2)x=0$
Prove
dydx=11+x2
Solution:
Here,$y=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) $
Using sec−1
$x=cos−1⁡(1x)y=tan−1⁡(2x1−x2)+cos−1⁡(1−x21+x2)$
Put x=tan⁡θ$y=tan−1⁡(2tan⁡θ1−tan2⁡θ)+cos−1⁡(1−tan2⁡θ1+tan2⁡θ)y=tan−1⁡(tan⁡2θ)+cos−1⁡(cos⁡2θ)−(1)$

Using,

tan⁡2θ=2tan⁡θ1−tan2⁡θ

$1−tan2⁡θ1+tan2⁡θ=cos⁡2θ Here, 0<x<∞0<tan⁡θ<∞0<θ<π20<2θ<π$

So From eq(i)

y=2θ+2θ

{ Since, tan−1⁡(tan⁡θ)=θ if θ∈[−π2,π2] and cos−1⁡(cos⁡θ)=θ if θ[0,π]}

y=4θ

y=4tan−1⁡x [Usingx=tan⁡θ,θ=tan−1⁡x]

Differentiating it with respect t0 x

$dydx=41+x2⇒41+x2 Using d(tan−1⁡x)dx=11+x2$

Differentiation Exercise 10.3 Question 40

Answer: dydx=0
Hint:
ddx(xn)=nxn−1;ddx( constant )=0
Given:
y=sec−1⁡(x+1x−1)+sin−1⁡(x−1x+1)x>0
Solution:
$y=sec−1⁡(x+1x−1)+sin−1⁡(x−1x+1)x>0y=cos−1⁡(x−1x+1)+sin−1⁡(x−1x+1)$
Using sec−1⁡x=cos⁡(1x)
Since,cos−1⁡x+sin−1⁡x=π2
y=π2
Differentiating with respect to x
dydx=0 [ddx(constant)=a]]

Differentiation Exercise 10.3 Question 41


Answer:dydx=−x1−x2
Hint:
$ddx( constant )=0;ddx(xn)=nxn−1$
Given:
y=sin⁡[2tan−1⁡{x1+x}]
Solution:
Put x=cos⁡2θ
$y=sin⁡[2tan−1⁡1−cos⁡2θ1+cos⁡2θ]y=sin⁡[2tan−1⁡2sin2⁡θ2cos2⁡θ] Since 1−cos⁡2θ=2sin2⁡θ1+cos⁡2θ=2cos2⁡θ$
$1+cos⁡2θ=2cos2⁡θy=sin⁡[2tan−1⁡tan2⁡θ] Since sin⁡θcos⁡θ=tan⁡θy=sin⁡[2tan−1⁡(tan⁡θ)]y=sin⁡[2θ]$ {tan−1⁡(tan⁡θ)=θifθε[−π2,π2]}
y=sin⁡(2×12cos−1⁡x) [Since,x=cos⁡2θ,θ=12cos−1⁡x]
$y=sin⁡(cos−1⁡x)y=sin⁡(sin−1⁡1−x2)y=1−x2$ [sin−1⁡(sin⁡θ)⇒0 if θ∈[−π2,π2]]
Differentiating it with respect to x
$dydx=ddx(1−x2)=ddx(1−x2)1/2$
As we know,
$Missing \end{aligned} \end{aligned}$

Differentiation Exercise 10.3 Question 42

Answer: dydx=21−4x2
Hint:
$ddx (constant) =0ddx(xn)=nx−1$
Given:
y=cos−1⁡(2x)+2cos−1⁡1+4x2,0<x<12
Solution:
Put 2x=cos⁡θ
So,
$y=cos−1⁡(cos⁡θ)+2cos−1⁡1−cos2⁡θ Since cos2⁡θ+sin2⁡θ=1y=cos−1⁡(cos⁡θ)+2cos−1⁡sin2⁡θy=cos−1⁡(cos⁡θ)+2cos−1⁡(sin⁡θ)y=cos⁡(cos⁡θ)+2cos−1⁡(cos⁡(π2−θ))−(i)$
$0<x=10<2x<10<cos⁡θ=10<θ<π2 And 0>−θ>−π2n2>(n2−θ)>0$
So From eq (i)
y=θ+2(π2−θ) {Sincecos−1⁡(cos⁡θ)=0ifθε[0,n]}
$y⇒θ+π−2θy⇒π−Cy=π−cos−1⁡(2x) Since, 2x=cos⁡θ$
Differentiating it with respect to x
dydx=0−[−11−(2x)2]ddx(2x)
Since,ddx(constant)=0;
$ddx(cos−1⁡x)=11−x2dydx=11−4x2(2)dydx=21−4x2$

Differentiation Exercise 10.3 Question 43

Answer: Hence Prove ,1+a2=b
Hint:
$ddx(xn)=nxn−1d( constant) dx=0$
Given:
ddx[tan−1⁡(a+bx)]=1 at x=0
Solution:
So, using chain rule,
we know ddx(tan−1⁡x)=11+x2
$[{11+(a+bx)2}ddx(a+bx)]x=0=1[11+(a+bx)2×(b)]x=0=1[b[1+(a+bx)2]x=0=1∴(a+b)2=a2+b2+2ab$
$[b1+(a2+b2x2+2abx)]x=0=1b1+(a2+0+0)=1b1+a2=1b=1+a2$
Hence Proved,

Differentiation Exercise 10.3 Question 44

Answer:dydx=−61−4x2
Hint:
$ddx(xn)=nxn−1ddx( Constant )=0$
Given:

Solution:
$y=cos−1⁡(2x)+2cos−1⁡1−4x2,−12<x<0$
Put2x=cos⁡θ
So
y=cos−1⁡(cocθ)+2cos−1⁡1−cos2⁡θ
Using,sin2⁡θ+cos2⁡θ=1
$y=cos−1⁡(cos⁡θ)+2cos⁡sin2⁡θy=cos−1⁡(cos⁡θ)+2cos−⁡(sin⁡θ)y=cos−1⁡(cos⁡θ)+2cos⁡(cos−1⁡(π2−θ))−(i)$
Considering limit,
$−12<x<0−1<2x<0−1−cos⁡θ<0π2<θ<π$
And,
$−π2>−θ>−π(π2−π2)>(π2−θ)>(π2−π)0>(π2−θ)>−π2$
So, from equation (i)

y=θ+2[−(π2−θ)]
{Since,cos−1⁡(cos⁡θ)=0 if θε[0,π],cos−1⁡cos⁡(θ)=−θ, if θε[−π⋅0]}$y=θ−2×π2+2θy=−π+3θy=−π+3cos−1⁡(2x)$

Differentiating its with respect to xusing chain rule

dydx=0+3[−11−(2x)2]ddx(2x)

As we Know,

$ddx( constant )=0ddx(cos−1⁡x)=−11−x2dydx=−31−4x2−(2)dydx=−61−4x2$

Differentiation Exercise 10.3 Question 45

Answer:dydx=121−x3
Hint:
$ddx( constant )=0ddx(xn)=nxn−1$
Given:
y=tan−1⁡{1+x−1−x1+x+1−x}
Solution:
Put,
x=cos⁡2θ
So,
$y=tan−1⁡[1+cos⁡2θ−1−cos⁡2θ1+cos⁡2θ+1−cos⁡2θ]=tan−1⁡(2cos2⁡θ−2sin2⁡θ2cos2⁡θ+2sin2⁡θ)$
Using$1+cos⁡2θ=2cos2⁡θy=tan−1⁡(2cos⁡θ−2sin⁡θ2cos⁡θ+2sin⁡θ)y=tan−1⁡(2(cos⁡θ−sin⁡θ)2(cos⁡θ+sin⁡θ))$

Dividing numerator and denominator by cos⁡θ

$y=tan−1⁡(cos⁡θcos⁡θ−sin⁡θcos⁡θcos⁡θcos⁡θ+sin⁡θcos⁡θ)y=tan−1⁡(cos⁡θ−sin⁡θcos⁡θcos⁡θ+sin⁡θcos⁡θ)$

Using

$tan⁡θ=sin⁡θcos⁡θy=tan−1⁡(1−tan⁡θ1+tan⁡θ)y=tan−1⁡(tan⁡n4−tan⁡θ1+tan⁡π4tan⁡θ)y=tan−1⁡(tan⁡(n4−θ)) Using tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B$

y=π−04−θ [Usingtan−1⁡(tan⁡θ)=θifθε[−π2,π2]]

y=π4−12cos−1⁡x[ Using x=cos⁡2θ]
Differentiating it with respect to x
dydx=0−12(−11−x2){ddx(cos−1⁡x)=−11−x2}
$dydx=121−x2dydx=−11−x2$

Differentiation Exercise 10.3 Question 47

Answer: dydx=−11−x2
Hint:
$ddx( constsant )=0ddx(xn)=nxn−1$
Given:
y=cos−1⁡{2x−31−x213)
Solution:
Let,x=cos⁡θ
$y=cos−1⁡{2cos⁡θ−31−cos2⁡θ4} Using cos2⁡θ+sin2⁡θ=1y=cos−1⁡{213cos⁡θ−313sin2⁡θ]y=cos−1⁡{213cos⁡θ−313sin⁡θ]$
cos⁡ϕ=213
Let,sin⁡ϕ=1−cos2⁡ϕ
$=1−(213)2+1−413=13−413=913$
sin⁡ϕ=313
So,
$Missing \end{aligned} \ &\text { Using, } \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned}$
y=ϕ+θ[cos−1⁡(cos⁡θ)=θ if 0ε[0,π]]
$y=cos−1⁡(213)+cos−1⁡x since, x=cos⁡θcos⁡ϕ=213$
Differentiating its with Respect to x,
$dydx=−11−(213)2ddx(213)+(−11−x2)dydx=−11−(213)2×0+(−11−x2)ddx(cos−1⁡x)=−11−x2dydx=0−11−x2$

Differentiation Exercise 10.3 Question 48

Answer: dydx=61−9x2
Hint:
$ddx (constant )=0ddx(xn)=nxn−1$
Given:
$y=sin−1⁡(6x1−9x2)132<x<132$
Solution:
Let
$x=13sin⁡θy=sin−1⁡(6×13sin⁡θ1−9(13)2sin2⁡θ)y=sin−1⁡(2sin⁡θ1−9×19sin2⁡θ)$
$y=sin−1⁡(2sin⁡θ1−sin2⁡θ)Usin⁡gsin2⁡θ+cos2⁡θ=1y=sin−1⁡(2sin⁡θcos2⁡θ)y=sin−1⁡(2sin⁡θcos⁡θ)y=sin−1⁡(sin⁡2θ)−(i)Usin⁡gsin⁡2θ=2sin⁡θcos⁡θ$
Considering limits here
$132<x<13213×132<13sin⁡θ<132×13192<13sin⁡θ<192$

From Equation (i)

y=2θ $\left\{sin−1⁡(sin⁡θ)=θ if θε[−π2,π2]\right\}$

y=2sin−1⁡(3x)

Differentiating it with respect tox

$dydx=ddx(2sin−1⁡3x) As we thow, ddx(sin−1⁡x)=11−x2dydx=2×11−(3x)2ddx(3x)dydx=21−9x2×3dydx=61−9x2$



Rd Sharma Class 12th Exercise 10.3 solutions are an ideal choice for every student because of the following benefits:

  1. Experts format this exclusive material

Rd Sharma Class 12th Exercise 10.3 solutions are created by a group of experts working day and night to make sure that every student understands the concept properly. Therefore, a student will always be confident about the exam after going for Rd Sharma Class 12 Chapter 10.3 Exercise 10.3 solutions.

  1. Best Solutions for preparation and revision

NEET Highest Scoring Chapters & Topics
This ebook serves as a valuable study guide for NEET exams, specifically designed to assist students in light of recent changes and the removal of certain topics from the NEET exam.
Download E-book

Rd Sharma Class 12th Exercise 10.3 solutions are the best source for preparation and revision because a lot of time is saved when a student studies from this material. Moreover, the students will be ready to face any exam with RD Sharma Class 12 Solutions Differentiation Ex. 10.3.

  1. Questions from the NCERT book are easy to understand:

It is a well-known fact that NCERT is a book that every teacher follows. Keeping this fact in mind, the students are provided with solutions that abide by NCERT and CBSE guidelines. Rd Sharma Class 12th Exercise 10.3 solutions help students achieve good marks in-class exams also.

  1. Alternative ways to solve a question

RD Sharma Class 12 Solutions Chapter 10 ex 10.3 because they are crafted by experts who have many ways to solve a single question. Every expert has a different way to solve a problem, here also a student will find many forms and can choose what suits them best.

  1. Great performance

These solutions help a student cross their benchmark scores in the exams. The Class 12th RD Sharma Chapter 10.3 Exercise 10.3 Solutions includes all the concepts that are present in the textbook.

  1. Free of cost

The students need not pay any kind of monetary fund to attain the book from the Career 360 website. This is the website that has benefited thousands of students all over the country. A student must follow this website and learn from Career360 to score well academically.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. What do you mean by Function Differentiation?

The Differentiation of a function is a mathematical operation in the calculus domain that identifies the instantaneous changes for a universal output.

2. Why should we learn the rule for calculating exponential functions?

The rule for deriving exponential functions should be learned to understand how the exponential expressions in a function change when a differentiation operation is performed. In addition, it will assist you in comprehending the advanced concepts of higher mathematics in higher classes

3. What is the minimum cost of the RD Sharma solution books?

The RD Sharma solution books are available for free of cost at the Career 360 website. The class 12 students can download the reference material to make the most of it. 

4. What are the different rules of Derivatives?

There are five types of rules of derivatives, and these are:

  • Power rule of derivatives

  • Sum rule of derivatives

  • Product rule of derivatives

  • Quotient rule of derivatives

  • Chain rule of derivatives

5. What Is the Best Way to Learn Derivative Functions?

If you follow Career360, you will quickly learn how to perform this set of actions on the functions. Practicing and learning are essential to apply such derivation rules easily.

Articles

Upcoming School Exams

Telangana Residential Junior Colleges Common Entrance Test

Application Date:24 March,2025 - 23 April,2025

National Institute of Open Schooling 10th examination

Admit Card Date:02 April,2025 - 19 May,2025

National Institute of Open Schooling 12th Examination

Admit Card Date:02 April,2025 - 19 May,2025

Telangana Open School Society Intermediate Examination

Admit Card Date:04 April,2025 - 26 April,2025

Telangana Open School Society SSC Examination

Admit Card Date:04 April,2025 - 26 April,2025

View All School Exams

Also Read

Study Resources, Applications and Opportunities

NEET/JEE Coaching Scholarship
NEET/JEE Coaching Scholarship
Apply

Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes

NEET Most scoring concepts
NEET Most scoring concepts
Get now

This ebook serves as a valuable study guide for NEET 2025 exam.

NEET Previous 10 Year Questions
NEET Previous 10 Year Questions
Get now

This e-book offers NEET PYQ and serves as an indispensable NEET study material.

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Get now

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
Get now

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

View All Application Forms
News and Notifications
Maharashtra govt stays order to make Hindi mandatory as 3rd language in Classes 1 to 5
Maharashtra won’t fully adopt CBSE curriculum, Hindi not compulsory: Dadaji Bhuse clarifies
Dharmendra Pradhan directs NCERT to introduce road safety modules in curriculum of students of Class 1-12
School jobs: HC to hear contempt plea against WB School Education Department for not uploading OMR sheets
Maharashtra govt retreats as state panel chief opposes mandatory Hindi from Class 1
Jobless teachers rally outside West Bengal SSC headquarters, demand release of 2016 recruitment list
West Bengal Recruitment Scam: Calcutta HC to hear primary teachers' termination appeals on April 28
Back to top