Question

Find the derivative $\dfrac{{dy}}{{dx}}$ for a given function $y = {\sin ^{ - 1}}(\dfrac{{1 - {x^2}}}{{1 + {x^2}}})$

Answer 1

Hint: Derivative of the inverse function can be easily obtained by using the standard formulas. Before using the standard formula, try to convert the function in the easiest form. Function $y = {\sin ^{ - 1}}(\dfrac{{1 - {x^2}}}{{1 + {x^2}}})$ can be simplified by assuming $x = \tan \theta $. And put $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and use standard formula $\operatorname{Cos} (2\theta ) = {\cos ^2}\theta - {\sin ^2}\theta $ and ${\sin ^2}\theta + {\cos ^2}\theta = 1$. After that use $\cos \theta = \sin ((4n + 1)\dfrac{\pi }{2} - \theta )$ to convert in $\sin \theta $ form. After that use the standard derivation formula of ${\tan ^{ - 1}}x$ to get $\dfrac{{dy}}{{dx}}$.
Standard formula for derivation of ${\tan ^{ - 1}}x$ is $\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = \dfrac{1}{{1 + {x^2}}}$.

Complete step-by-step answer:
Here given function, $y = {\sin ^{ - 1}}(\dfrac{{1 - {x^2}}}{{1 + {x^2}}})$
Before finding derivatives of function, let’s simplify it.
Let’s assume $x = \tan \theta $
So, $y = {\sin ^{ - 1}}(\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }})$
Putting $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
So, $y = {\sin ^{ - 1}}(\dfrac{{1 - \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{1 + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}})$
Simplifying the equitation,
 \[y = {\sin ^{ - 1}}(\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta + {{\sin }^2}\theta }})\]
Using formula $\operatorname{Cos} (2\theta ) = {\cos ^2}\theta - {\sin ^2}\theta $ and ${\sin ^2}\theta + {\cos ^2}\theta = 1$
So, \[y = {\sin ^{ - 1}}(\dfrac{{\cos (2\theta )}}{1})\]
\[y = {\sin ^{ - 1}}(\cos (2\theta ))\]
Now converting $\cos \theta $ function in $\sin \theta $ function by using formula, $\cos \theta = \sin ((4n + 1)\dfrac{\pi }{2} - \theta )$, where n is any natural number ($n \in N$).
So, converting the above equitation,
\[y = {\sin ^{ - 1}}(\sin ((4n + 1)\dfrac{\pi }{2} - 2\theta ))\]
As we know that, \[{\sin ^{ - 1}}(\sin \theta ) = \theta \], so simplifying above equation,
\[y = (4n + 1)\dfrac{\pi }{2} - 2\theta )\]
Now as we assumed $x = \tan \theta $, so $\theta = {\tan ^{ - 1}}x$
Now putting $\theta = {\tan ^{ - 1}}x$ in above equation,
\[y = (4n + 1)\dfrac{\pi }{2} - 2{\tan ^{ - 1}}x\]
Above equation is the simplified form of the given function $y = {\sin ^{ - 1}}(\dfrac{{1 - {x^2}}}{{1 + {x^2}}})$.
Now derivations of the function $y$ can be obtained by derivation with respect to $x$.
So, derivation of function $y$ with respect to $x$, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}((4n + 1)\dfrac{\pi }{2} - 2{\tan ^{ - 1}}x)$
Using basic rule of derivatives of sum function, $\dfrac{d}{{dx}}(f(x) + g(x)) = \dfrac{d}{{dx}}(f(x)) + \dfrac{d}{{dx}}(g(x))$,
Simplifying our equation $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}((4n + 1)\dfrac{\pi }{2}) + \dfrac{d}{{dx}}( - 2{\tan ^{ - 1}}x)$
As $\dfrac{d}{{dx}}(c) = 0$ means derivation of any constant term is zero, so, $\,\dfrac{d}{{dx}}((4n + 1)\dfrac{\pi }{2}) = 0$.
So, $\dfrac{{dy}}{{dx}} = 0 + \dfrac{d}{{dx}}( - 2{\tan ^{ - 1}}x)$
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}( - 2{\tan ^{ - 1}}x)$
Using $\dfrac{d}{{dx}}(af(x)) = a\dfrac{d}{{dx}}(f(x)$ in our equation, we will get $\dfrac{{dy}}{{dx}} = ( - 2)\dfrac{d}{{dx}}({\tan ^{ - 1}}x)$
Using standard formula for derivation of ${\tan ^{ - 1}}x$, $\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = \dfrac{1}{{1 + {x^2}}}$
Putting this in our above equation,
$\dfrac{{dy}}{{dx}} = ( - 2) \times (\dfrac{1}{{1 + {x^2}}})$
So, $\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{1 + {x^2}}}$

So, derivation of given function $y = {\sin ^{ - 1}}(\dfrac{{1 - {x^2}}}{{1 + {x^2}}})$ is $\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{1 + {x^2}}}$

Note: If same type of function $y = {\sin ^{ - 1}}(\dfrac{{2x}}{{1 + {x^2}}})$ is given then also assume $x = \tan \theta $ and use $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ to simplify.
After that use ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $\sin (2\theta ) = 2\sin \theta \cos \theta $ to convert $\dfrac{{2x}}{{1 + {x^2}}} = \sin 2\theta $, So \[y = {\sin ^{ - 1}}(\sin (2\theta )) = 2\theta \].
Then put $\theta = {\tan ^{ - 1}}x$, so \[y = 2{\tan ^{ - 1}}x\].
After that do derivation of the function $y$.
So $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(2{\tan ^{ - 1}}x) = 2\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = 2(\dfrac{1}{{1 + {x^2}}}) = \dfrac{2}{{1 + {x^2}}}$.