Question

Differentiate the following with respect to x:
${{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}}$,$\dfrac{{1}}{\sqrt{2}} $ < x < 1

Answer 1

- Hint: We will be using the concept of inverse trigonometric function to simplify the expression and then we will be using the concepts of differential calculus.

Complete step-by-step solution -

Now, we have been given a function $f\left( x \right)={{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}},\dfrac{1}{\sqrt{2}} < x < 1$.
We have to find the $\dfrac{d}{dx}f\left( x \right)$ or we have to find the derivative of the function.
We will first simplify the $f\left( x \right)$ for this. Let us take $x=\sin \theta \ $since $-1\le \sin \theta \le 1\ $and also $\dfrac{1}{\sqrt{2}} < x < 1$. Therefore it can fit x easily. Now, we have $f\left( x \right)$ as
$={{\cos }^{-1}}2\sin \theta \sqrt{1-{{\sin }^{2}}\theta }$
Now, we know that,
$\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
\end{align}$
So, we will use this to replace the value of $1-{{\sin }^{2}}\theta $
$\begin{align}
  & ={{\cos }^{-1}}\left( 2\sin \theta \sqrt{{{\cos }^{2}}\theta } \right) \\
 & ={{\cos }^{-1}}\left( 2\sin \theta \cos \theta \right) \\
\end{align}$
Now, we know that $\sin 2\theta =2\sin \theta \cos \theta $,
So, we have,
$={{\cos }^{-1}}\left( \sin 2\theta \right)$
Also, we know that,
$\begin{align}
  & \cos \left( \dfrac{\pi }{2}-20 \right)=\sin 2\theta \\
 & \therefore {{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-2\theta \right) \right) \\
\end{align}$
We know that ${{\cos }^{-1}}\left( \cos x \right)=x$.
So, using this we have,
${{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-2\theta \right) \right)=\dfrac{\pi }{2}-2\theta ..........\left( 1 \right)$
Now, we have taken $x=\sin 2\theta $. So, we will find the value of $\theta $ from it and substitute in (1).
$\begin{align}
  & {{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}}=\dfrac{\pi }{2}-2\theta =\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\
 & f\left( x \right)=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\
\end{align}$
Now, we differentiate $f\left( x \right)$ with the respect to x, to get the answer.
We know that,
$\dfrac{d}{dx}{{\sin }^{-1}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$
Therefore,
$\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{\pi }{2}-2{{\sin }^{-1}}x \right) \\
 & =-2\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right) \\
 & =-2\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}} \right) \\
 & =\dfrac{-2}{\sqrt{1-{{x}^{2}}}} \\
\end{align}$
So, the differentiation of ${{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}}$ is $\dfrac{-2}{\sqrt{1-{{x}^{2}}}}$.

Note: To solve these types of questions it is important to note that we have used trigonometric identity that $\sin 2\theta =2\sin \theta \cos \theta $ to simplify the inverse trigonometric function and then we have used the concept of differential calculus to find the final answer.