Question

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

Answer 1

Hint: 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 :
\[{\sin ^{ - 1}}\left( { - x} \right) = - {\sin ^{ - 1}}x\] , \[x \in \left[ { - 1,1} \right]\]
\[{\cos ^{ - 1}}\left( { - x} \right) = \pi - {\cos ^{ - 1}}x\] , \[x \in \left[ { - 1,1} \right]\]
\[{\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}\left( x \right)\] , \[x \in R\]