मराठी

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

प्रश्न

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

बेरीज
Advertisements

उत्तर

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

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

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

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

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

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

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

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

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

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

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

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

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

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.08 [पृष्ठ ११३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.08 | Q 12 | पृष्ठ ११३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is `cos^(-1)(1/sqrt3)`


Prove that `y=(4sintheta)/(2+costheta)-theta `


Differentiate tan 5x° ?


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


Differentiate (log sin x)?


Differentiate \[e^\sqrt{\cot x}\] ?


Differentiate \[\frac{e^x \log x}{x^2}\] ? 


Differentiate \[e^{ax} \sec x \tan 2x\] ?


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


Differentiate \[\tan^{- 1} \left( \frac{a + b \tan x}{b - a \tan x} \right)\] ?


Find  \[\frac{dy}{dx}\] in the following case \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] ?


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


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


Differentiate \[\left( \log x \right)^x\] ?


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


Differentiate \[e^{\sin x }+ \left( \tan x \right)^x\] ?


Find \[\frac{dy}{dx}\]  \[y = x^n + n^x + x^x + n^n\] ?

Find  \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?

 


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


If \[x^y + y^x = \left( x + y \right)^{x + y} , \text{ find } \frac{dy}{dx}\] ?


If \[y = \left( \cos x \right)^{\left( \cos x \right)^{\left( \cos x \right) . . . \infty}}\],prove that \[\frac{dy}{dx} = - \frac{y^2 \tan x}{\left( 1 - y \log \cos x \right)}\]?

 


If \[\frac{dy}{dx}\] when \[x = a \cos \theta \text{ and } y = b \sin \theta\] ?


\[\sin x = \frac{2t}{1 + t^2}, \tan y = \frac{2t}{1 - t^2}, \text { find }  \frac{dy}{dx}\] ?

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


The differential coefficient of f (log x) w.r.t. x, where f (x) = log x is ___________ .


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


If \[f\left( x \right) = \sqrt{x^2 + 6x + 9}, \text { then } f'\left( x \right)\] is equal to ______________ .


If \[y = \frac{1}{1 + x^{a - b} +^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}\] then \[\frac{dy}{dx}\]  is equal to ______________ .


If  \[\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3 \left( x^3 - y^3 \right)\] then \[\frac{dy}{dx}\] is equal to ____________ .


If \[y = \log \sqrt{\tan x}\] then the value of \[\frac{dy}{dx}\text { at }x = \frac{\pi}{4}\] is given by __________ .


Find the second order derivatives of the following function ex sin 5x  ?


If y = log (sin x), prove that \[\frac{d^3 y}{d x^3} = 2 \cos \ x \ {cosec}^3 x\] ?


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


If y = (tan−1 x)2, then prove that (1 + x2)2 y2 + 2x(1 + x2)y1 = 2 ?


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

 


If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to

 


If \[y = \frac{ax + b}{x^2 + c}\] then (2xy1 + y)y3 = 

 


Differentiate sin(log sin x) ?


Show that the height of a cylinder, which is open at the top, having a given surface area and greatest volume, is equal to the radius of its base. 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×