Question

Find the derivative of the following:
$y=\sin \left( {{\tan }^{-1}}x \right)$.

Answer 1

Hint: Assume $\sin x=f(x)$ and ${{\tan }^{-1}}x=g(x)$, so that, we get the function of the form $f\left( g(x) \right)$. To find the derivative of the function of the form: $f\left( g(x) \right)$, use the formula: $\dfrac{d}{dx}\left[ f\left( g(x) \right) \right]={{f}^{'}}\left( g(x) \right)\times \dfrac{d}{dx}\left[ g(x) \right]$. Use the derivative of $\sin x$ equal to $\cos x$ and the derivative of ${{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$.

Complete step-by-step answer:
Let us assume $\sin x=f(x)$ and ${{\tan }^{-1}}x=g(x)$, so that, we get the function of the form $f\left( g(x) \right)$. $f\left( g(x) \right)$ is a composite function.
A composite function is a function that depends on another function. It is created when one function is substituted into another function.
Now, to find the derivative of a function of the form $f\left( g(x) \right)$, we use the formula given by: $\dfrac{d}{dx}\left[ f\left( g(x) \right) \right]={{f}^{'}}\left( g(x) \right)\times \dfrac{d}{dx}\left[ g(x) \right]$.
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left[ f\left( g(x) \right) \right] \\
 & ={{f}^{'}}\left( g(x) \right)\times \dfrac{d}{dx}\left[ g(x) \right] \\
 & =\dfrac{d}{dx}\left[ \sin \left( {{\tan }^{-1}}x \right) \right] \\
 & =\cos \left( {{\tan }^{-1}}x \right)\times \dfrac{d}{dx}\left[ {{\tan }^{-1}}x \right] \\
\end{align}$
We know that the derivative of ${{\tan }^{-1}}x$ is $\dfrac{1}{1+{{x}^{2}}}$. Therefore,
$\dfrac{d}{dx}\left[ \sin \left( {{\tan }^{-1}}x \right) \right]=\cos \left( {{\tan }^{-1}}x \right)\times \dfrac{1}{1+{{x}^{2}}}$

Note: You may note that we have written the given function in the form of a composite function. The reason behind this is that we have to apply the derivative formula: $\dfrac{d}{dx}\left[ f\left( g(x) \right) \right]={{f}^{'}}\left( g(x) \right)\times \dfrac{d}{dx}\left[ g(x) \right]$. We can also solve this question by an alternate method. In this alternate method, we have to change the given ${{\tan }^{-1}}x$ into sine inverse function by assuming ‘x’ as the perpendicular ‘1’ as the base. Once it is converted to a sine inverse function, apply the formula: $\sin \left( {{\sin }^{-1}}x \right)=x$, so that we are left with only an algebraic function. Now, we can differentiate this algebraic function easily.