English

Differentiate Tan − 1 ( 2 X 1 − X 2 ) with Respect to Cos − 1 ( 1 − X 2 1 + X 2 ) , If 0 < X < 1 ? - Mathematics

Advertisements
Advertisements

Question

Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text {  if }0 < x < 1\] ?

Sum
Advertisements

Solution

\[\text { Let, u } = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]

\[\text { Put x } = \tan\theta\]

\[ \Rightarrow u = \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right)\]

\[ \Rightarrow u = \tan^{- 1} \left( \tan2\theta \right) . . . \left( i \right)\]

\[\text { let, v} = \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)\]

\[ \Rightarrow v = \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \]

\[ \Rightarrow v = \cos^{- 1} \left( \cos2\theta \right) . . . \left( ii \right)\]

\[\text { Here, } 0 < x < 1\]

\[ \Rightarrow 0 < \tan\theta < 1\]

\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]

\[\text { So, from equation } \left( i \right), \]

\[u = 2\theta \left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta,\text { if } \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]

\[ \Rightarrow u = 2 \tan^{- 1} x \left[ \text { Since,} x = \tan\theta \right]\]

differentiating it with respect to x,

\[\frac{du}{dx} = \frac{2}{1 + x^2} . . . \left( iii \right)\]

\[\text { From equation } \left( ii \right), \]

\[v = \theta .........\left[ \text { Since }, \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right]\]

\[ \Rightarrow v = 2 \tan^{- 1} x ......\left[ \text { Since, } x = \tan\theta \right]\]

Differentiating it with respect to x,

\[\frac{dv}{dx} = \frac{2}{1 + x^2} . . . \left( iv \right)\]

\[\text { Dividing equation } \left( iii \right) \text { by }\left( iv \right), \]

\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2}\]

\[ \therefore \frac{du}{dv} = 1\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Differentiation - Exercise 11.08 [Page 113]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 11 Differentiation
Exercise 11.08 | Q 12 | Page 113

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

 

If y = xx, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`

 

Differentiate sin (3x + 5) ?


Differentiate sin (log x) ?


Differentiate \[e^{\sin} \sqrt{x}\] ?


Differentiate \[\log \left( x + \sqrt{x^2 + 1} \right)\] ?


\[\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}\] ?


If \[y = x \sin^{- 1} x + \sqrt{1 - x^2}\] ,prove that \[\frac{dy}{dx} = \sin^{- 1} x\] ?


If \[y = \sqrt{x^2 + a^2}\] prove that  \[y\frac{dy}{dx} - x = 0\] ?


If \[y = \sqrt{a^2 - x^2}\] prove that  \[y\frac{dy}{dx} + x = 0\] ?


Prove that \[\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}\] ?


Differentiate \[\cos^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?


If \[y = \sin^{- 1} \left( \frac{x}{1 + x^2} \right) + \cos^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right), 0 < x < \infty\] prove that  \[\frac{dy}{dx} = \frac{2}{1 + x^2} \] ?

 


If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, - \frac{1}{2} < x < 0, \text{ find } \frac{dy}{dx} \] ?


If \[\sin^2 y + \cos xy = k,\] find  \[\frac{dy}{dx}\] at \[x = 1 , \] \[y = \frac{\pi}{4} .\] 


Differentiate \[{10}^{ \log \sin x }\] ?


Differentiate  \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\]  ?


If `y=(sinx)^x + sin^-1 sqrtx  "then find"  dy/dx` 


If  \[\left( \sin x \right)^y = x + y\] , prove that \[\frac{dy}{dx} = \frac{1 - \left( x + y \right) y \cot x}{\left( x + y \right) \log \sin x - 1}\] ?

 


If \[y = e^{x^{e^x}} + x^{e^{e^x}} + e^{x^{x^e}}\], prove that  \[\frac{dy}{dx} = e^{x^{e^x}} \cdot x^{e^x} \left\{ \frac{e^x}{x} + e^x \cdot \log x \right\}+ x^{e^{e^x}} \cdot e^{e^x} \left\{ \frac{1}{x} + e^x \cdot \log x \right\} + e^{x^{x^e}} x^{x^e} \cdot x^{e - 1} \left\{ x + e \log x \right\}\]

 


Find \[\frac{dy}{dx}\] ,When \[x = e^\theta \left( \theta + \frac{1}{\theta} \right) \text{ and } y = e^{- \theta} \left( \theta - \frac{1}{\theta} \right)\] ?


Differentiate x2 with respect to x3


Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( \frac{1}{2 \sqrt{2}}, \frac{1}{2} \right)\] ?


Differentiate\[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)\] with respect to \[\sin^{-1} \left( \frac{2x}{1 + x^2} \right)\], If \[- 1 < x < 1, x \neq 0 .\] ?


Differentiate \[\left( \cos x \right)^{\sin x }\] with respect to \[\left( \sin x \right)^{\cos x }\]?


Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right), \text { if }- \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\] ?


Differentiate \[\sin^{- 1} \left( 2 ax \sqrt{1 - a^2 x^2} \right)\] with respect to \[\sqrt{1 - a^2 x^2}, \text{ if }-\frac{1}{\sqrt{2}} < ax < \frac{1}{\sqrt{2}}\] ?


If \[f'\left( x \right) = \sqrt{2 x^2 - 1} \text { and y } = f \left( x^2 \right)\] then find \[\frac{dy}{dx} \text { at } x = 1\] ?


If \[f\left( x \right) = \log \left\{ \frac{u \left( x \right)}{v \left( x \right)} \right\}, u \left( 1 \right) = v \left( 1 \right) \text{ and }u' \left( 1 \right) = v' \left( 1 \right) = 2\] , then find the value of `f' (1)` ?


If \[y = \log \left| 3x \right|, x \neq 0, \text{ find } \frac{dy}{dx} \] ? 


Differential coefficient of sec(tan−1 x) is ______.


For the curve \[\sqrt{x} + \sqrt{y} = 1, \frac{dy}{dx}\text {  at } \left( 1/4, 1/4 \right)\text {  is }\] _____________ .


If \[3 \sin \left( xy \right) + 4 \cos \left( xy \right) = 5, \text { then } \frac{dy}{dx} =\] _____________ .


Find the second order derivatives of the following function x3 log ?


If \[y = e^{2x} \left( ax + b \right)\]  show that  \[y_2 - 4 y_1 + 4y = 0\] ?


If y = ex (sin + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?


If y = 3 e2x + 2 e3x, prove that  \[\frac{d^2 y}{d x^2} - 5\frac{dy}{dx} + 6y = 0\] ?


If y = x + ex, find \[\frac{d^2 x}{d y^2}\] ?


\[\frac{d^{20}}{d x^{20}} \left( 2 \cos x \cos 3 x \right) =\]

 


If y = etan x, then (cos2 x)y2 =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×