Question

Differentiate ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ with respect to ${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$ .

Answer 1

We have to find the derivative of the given trigonometric function ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ with respect to another trigonometric function ${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$ . We solve this question using the concept of differentiation of trigonometric functions and the various formulas for the inverse of trigonometric functions . We solve this question by substituting the value of the given function . First we will substitute the value of angle of both the trigonometric functions such that we get a simplified form of the trigonometric functions . Then we will differentiate the expression of both the functions with respect to $x$ . And finally we will divide the derivative of the tangent function by the derivative of the sine function to get the required answer of the derivative .

Complete step-by-step solution:
Given :
We have to differentiate ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ with respect to ${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$
Let $u = {\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ and $v = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$
Now , we have to find $\dfrac{{du}}{{dv}}$ .
For the value of $\dfrac{{du}}{{dv}}$ , we will differentiate both $u$ and $v$ separately with respect to $x$ .
Now , we will solve the value of the derivative in two cases: one for $u$ and second for $v$ .
$Case{\text{ }}1{\text{ }}:$
$u = {\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$
Put $x = \sin a$
Also , $a = {\sin ^{ - 1}}x$
Substituting the values , we get
$u = {\tan ^{ - 1}}\dfrac{{\sin a}}{{\sqrt {1 - {{\sin }^2}a} }}$
We also know that the relation of sine and cosine function is given as :
${sin^2}x + {cos^2}x = 1$
${cos^2}x = 1 - {sin^2}x$
Using the above expression , we get the value of $u$ as :
$u = {\tan ^{ - 1}}\dfrac{{\sin a}}{{\cos a}}$
We also know that :
$\tan u = \dfrac{{\sin a}}{{\cos a}}$
So , the expression of $u$ becomes :
$u = {\tan ^{ - 1}}\left( {\tan a} \right)$
Now , all know that :
${\tan ^{ - 1}}\left( {\tan a} \right) = a$ , $a \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$
Using the relation , we get value of $u$ as :
$u = a$
Now substituting the value of a back , we get the expression as :
$u = {\sin ^{ - 1}}x$
We also , know that the derivative of ${\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Using the formula , we get the value of $\dfrac{{du}}{{dx}}$ as :
Differentiating $u$ with respect to $x$ , we get
$\dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }} - - - \left( 1 \right)$
$Case{\text{ }}2{\text{ }}:$
$v = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$
Put $x = \sin b$
Also , $b = {\sin ^{ - 1}}x$
Substituting the values , we get
$v = {\sin ^{ - 1}}\left( {2\sin b\sqrt {1 - {{\sin }^2}b} } \right)$
We also know that the relation of sine and cosine function is given as :
${sin^2}x + {cos^2}x = 1$
${cos^2}x = 1 - {sin^2}x$
Using the above expression , we get the value of $v$ as :
$v = {\sin ^{ - 1}}\left( {2\sin b\cos b} \right)$
We also know that the formula of double angle of sine function is given as :
$\sin 2x = 2\sin x\cos x$
So , the expression of $v$ becomes :
$v = {\sin ^{ - 1}}\left( {\sin 2b} \right)$
Now , all know that :
${\sin ^{ - 1}}\left( {\sin x} \right) = x$ , $x \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$
Using the relation , we get value of $v$ as :
$v = 2b$
Now substituting the value of b back , we get the expression as :
$v = 2{\sin ^{ - 1}}x$
We also , know that the derivative of ${\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Using the formula , we get the value of $\dfrac{{dv}}{{dx}}$ as :
Differentiating $v$ with respect to $x$ , we get
$\dfrac{{dv}}{{dx}} = \dfrac{2}{{\sqrt {1 - {x^2}} }} - - - \left( 2 \right)$
Now , dividing the two equations , we get the value of $\dfrac{{du}}{{dv}}$ as :
$\dfrac{{du}}{{dv}} = \dfrac{{\left( {\dfrac{{du}}{{dx}}} \right)}}{{\left( {\dfrac{{dv}}{{dx}}} \right)}}$
$\dfrac{{du}}{{dv}} = \dfrac{{\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)}}{{\left( {\dfrac{2}{{\sqrt {1 - {x^2}} }}} \right)}}$
Cancelling the terms , we get
$\dfrac{{du}}{{dv}} = \dfrac{1}{2}$
Hence , the value of the differentiate ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {1 - {x^2}} }}$ with respect to ${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$ is $\dfrac{1}{2}$ .

Note: We used the concept of the range of the inverse of trigonometric functions . The range of the trigonometric function is stated as the value of the intervals between which all the values of the trigonometric function lie .
Various inverse trigonometric functions are given as :
${\sin ^{ - 1}}\left( { - x} \right) = - {\sin ^{ - 1}}x$ , $x \in \left[ { - 1,1} \right]$
${\cos ^{ - 1}}\left( { - x} \right) = \pi - {\cos ^{ - 1}}x$ , $x \in \left[ { - 1,1} \right]$
${\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}\left( x \right)$ , $x \in R$