Question

Differentiate the following with respect to $x$:
${\sin ^{ - 1}}\left( {\dfrac{{1 - 25{x^2}}}{{1 + 25{x^2}}}} \right)$

Answer 1

Hint: let $y = {\sin ^{ - 1}}\left( {\dfrac{{1 - 25{x^2}}}{{1 + 25{x^2}}}} \right)$ and $5x = \tan \theta $, then the expression will be $y = {\sin ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$. Simplify the expression using the properties of trigonometry. After simplifying the expression, put $\theta = {\tan ^{ - 1}}\left( {5x} \right)$ and apply chain rule to find the derivative of the given expression.

Complete step-by-step answer:
We will first let the given expression, ${\sin ^{ - 1}}\left( {\dfrac{{1 - 25{x^2}}}{{1 + 25{x^2}}}} \right)$ equals to $y$. We have to find the value of $\dfrac{{dy}}{{dx}}$
Since, we can see $25{x^2} = {\left( {5x} \right)^2}$, let $5x = \tan \theta $
Then,
$y = {\sin ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$
We also know that $\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) = \cos 2\theta $
Hence, $y = {\sin ^{ - 1}}\left( {\cos 2\theta } \right)$
Now, we can write $\cos \alpha = \sin \left( {\dfrac{\pi }{2} - \alpha } \right)$
$y = {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{2} - 2\theta } \right)} \right)$
Since, we have the property of inverse, that ${\sin ^{ - 1}}\left( {\sin x} \right) = x$
Then, $y = \dfrac{\pi }{2} - 2\theta $
Substitute back the value of $\theta $
We had let $5x = \tan \theta $, therefore, the value of $\theta = {\tan ^{ - 1}}\left( {5x} \right)$
$y = \dfrac{\pi }{2} - 2{\tan ^{ - 1}}\left( {5x} \right)$
Differentiate both sides with respect to $x$
Here, we will apply rain rule, which states that, $f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Also, it is known that $\dfrac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{1 + {x^2}}}$
$
  \dfrac{{dy}}{{dx}} = 0 - 2\dfrac{1}{{1 + {{\left( {5x} \right)}^2}}}\left( {\dfrac{{d\left( {5x} \right)}}{{dx}}} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 2\dfrac{1}{{1 + {{\left( {5x} \right)}^2}}}\left( 5 \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 10}}{{1 + 25{x^2}}} \\
$
Hence, the value of differentiation of ${\sin ^{ - 1}}\left( {\dfrac{{1 - 25{x^2}}}{{1 + 25{x^2}}}} \right)$ is $\dfrac{{ - 10}}{{1 + 25{x^2}}}$.

Note: Students must know the formulas of trigonometry to do this question correctly. Whenever we have the form $\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$, in the angle of an inverse function, we always substitute $x = \tan \theta $ and hence, form the formula of $\cos 2\theta $.