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Question: If \(y=\sin \left[ {{\cos }^{-1}}\left\\{ \sin \left( {{\cos }^{-1}}x \right) \right\\} \right]\), t...

If y=\sin \left[ {{\cos }^{-1}}\left\\{ \sin \left( {{\cos }^{-1}}x \right) \right\\} \right], then dydx\dfrac{dy}{dx} at x=12x=\dfrac{1}{2} is equal to?
(a) 00
(b) 11
(c) 23\dfrac{2}{\sqrt{3}}
(d) 13\dfrac{1}{\sqrt{3}}

Explanation

Solution

First of all simplify the given expression y. Use the formula cos⁡−1x+sin⁡−1x=π2{{\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⁡(π2−θ)=cos⁡θ\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta to simplify. Further, for the square bracket use the formula cos⁡−1(cos⁡θ)=θ{{\cos }^{-1}}\left( \cos \theta \right)=\theta for 0≤θ≤π0\le \theta \le \pi and finally use the formula sin⁡(sin⁡−1x)=x\sin \left( {{\sin }^{-1}}x \right)=x for −1≤x≤1-1\le x\le 1 to get the simplified expression of y. Differentiate both the sides with respect to x and substitute x=12x=\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 dydx\dfrac{dy}{dx} at x=12x=\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⁡−1x+sin⁡−1x=π2{{\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⁡(π2−θ)=cos⁡θ\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(cos⁡θ)=θ{{\cos }^{-1}}\left( \cos \theta \right)=\theta we must have the condition 0≤θ≤π0\le \theta \le \pi , where θ=sin⁡−1x\theta ={{\sin }^{-1}}x. We know that sin⁡−1x∈[−π2,π2]{{\sin }^{-1}}x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]. Since we have to find the value of dydx\dfrac{dy}{dx} at x=12x=\dfrac{1}{2} and around x=12x=\dfrac{1}{2} we can say that the inverse sine function lies in the range [0,π2]\left[ 0,\dfrac{\pi }{2} \right]. Therefore we have sin⁡−1x∈[0,π2]{{\sin }^{-1}}x\in \left[ 0,\dfrac{\pi }{2} \right] for the above case, so we can use the formula cos⁡−1(cos⁡θ)=θ{{\cos }^{-1}}\left( \cos \theta \right)=\theta .
⇒y=sin⁡[sin⁡−1x]\Rightarrow y=\sin \left[ {{\sin }^{-1}}x \right]
Finally, using the identity sin⁡(sin⁡−1x)=x\sin \left( {{\sin }^{-1}}x \right)=x for −1≤x≤1-1\le x\le 1 we get,
⇒y=x\Rightarrow y=x
Differentiating both the sides with respect to x we get,
⇒dydx=dxdx ∴dydx=1 \begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{dx}{dx} \\\ & \therefore \dfrac{dy}{dx}=1 \\\ \end{aligned}
Clearly dydx\dfrac{dy}{dx} is a constant term so for any value of x the value of dydx\dfrac{dy}{dx} is not going to change. Therefore, at x=12x=\dfrac{1}{2} we have dydx=1\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.