Question

If $y=\sin \left[ {{\cos }^{-1}}\left\{ \sin \left( {{\cos }^{-1}}x \right) \right\} \right]$, then $\dfrac{dy}{dx}$ at $x=\dfrac{1}{2}$ is equal to?
(a) $0$
(b) $1$
(c) $\dfrac{2}{\sqrt{3}}$
(d) $\dfrac{1}{\sqrt{3}}$

Answer 1

Hint: First of all simplify the given expression y. Use the formula ${{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2}$ to simplify the expression inside the small bracket. Now, come to the curly bracket and use the complementary angle formula given as $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ to simplify. Further, for the square bracket use the formula ${{\cos }^{-1}}\left( \cos \theta \right)=\theta $ for $0\le \theta \le \pi $ and finally use the formula $\sin \left( {{\sin }^{-1}}x \right)=x$ for $-1\le x\le 1$ to get the simplified expression of y. Differentiate both the sides with respect to x and substitute $x=\dfrac{1}{2}$ to get the answer.

Complete step-by-step solution:
Here we have been provided with the function $y=\sin \left[ {{\cos }^{-1}}\left\{ \sin \left( {{\cos }^{-1}}x \right) \right\} \right]$ and we are asked to find the value of $\dfrac{dy}{dx}$ at $x=\dfrac{1}{2}$. First we need to simplify the expression by using certain trigonometric and inverse trigonometric identities.
$\Rightarrow y=\sin \left[ {{\cos }^{-1}}\left\{ \sin \left( {{\cos }^{-1}}x \right) \right\} \right]$
Using the identity ${{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2}$ inside the small bracket we get,
$\Rightarrow y=\sin \left[ {{\cos }^{-1}}\left\{ \sin \left( \dfrac{\pi }{2}-{{\sin }^{-1}}x \right) \right\} \right]$
Using the complementary angle formula $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ we get,
$\Rightarrow y=\sin \left[ {{\cos }^{-1}}\left\{ \cos \left( {{\sin }^{-1}}x \right) \right\} \right]$
For using the formula ${{\cos }^{-1}}\left( \cos \theta \right)=\theta $ we must have the condition $0\le \theta \le \pi $, where $\theta ={{\sin }^{-1}}x$. We know that ${{\sin }^{-1}}x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Since we have to find the value of $\dfrac{dy}{dx}$ at $x=\dfrac{1}{2}$ and around $x=\dfrac{1}{2}$ we can say that the inverse sine function lies in the range $\left[ 0,\dfrac{\pi }{2} \right]$. Therefore we have ${{\sin }^{-1}}x\in \left[ 0,\dfrac{\pi }{2} \right]$ for the above case, so we can use the formula ${{\cos }^{-1}}\left( \cos \theta \right)=\theta $.
$\Rightarrow y=\sin \left[ {{\sin }^{-1}}x \right]$
Finally, using the identity $\sin \left( {{\sin }^{-1}}x \right)=x$ for $-1\le x\le 1$ we get,
$\Rightarrow y=x$
Differentiating both the sides with respect to x we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{dx}{dx} \\
 & \therefore \dfrac{dy}{dx}=1 \\
\end{align}$
Clearly $\dfrac{dy}{dx}$ is a constant term so for any value of x the value of $\dfrac{dy}{dx}$ is not going to change. Therefore, at $x=\dfrac{1}{2}$ we have $\dfrac{dy}{dx}=1$.


Note: We can also get the answer without simplifying and directly differentiating the function on both sides using the chain rule or derivative. However, in this case the calculation will be hard as there are two trigonometric and two inverse trigonometric functions that are to be differentiated. So it is necessary to simplify the function to the extent we can simplify.