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Question: Differentiate the following with respect to x: \({{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}}\),\(\dfrac{{1}...

Differentiate the following with respect to x:
cos⁡−12x1−x2{{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}},12\dfrac{{1}}{\sqrt{2}} < x < 1

Explanation

Solution

- 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(x)=cos⁡−12x1−x2,12<x<1f\left( x \right)={{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}},\dfrac{1}{\sqrt{2}} < x < 1.
We have to find the ddxf(x)\dfrac{d}{dx}f\left( x \right) or we have to find the derivative of the function.
We will first simplify the f(x)f\left( x \right) for this. Let us take x=sin⁡θ x=\sin \theta \ since −1≤sin⁡θ≤1 -1\le \sin \theta \le 1\ and also 12<x<1\dfrac{1}{\sqrt{2}} < x < 1. Therefore it can fit x easily. Now, we have f(x)f\left( x \right) as
=cos⁡−12sin⁡θ1−sin⁡2θ={{\cos }^{-1}}2\sin \theta \sqrt{1-{{\sin }^{2}}\theta }
Now, we know that,
sin⁡2θ+cos⁡2θ=1 cos⁡2θ=1−sin⁡2θ \begin{aligned} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\\ \end{aligned}
So, we will use this to replace the value of 1−sin⁡2θ1-{{\sin }^{2}}\theta
=cos⁡−1(2sin⁡θcos⁡2θ) =cos⁡−1(2sin⁡θcos⁡θ) \begin{aligned} & ={{\cos }^{-1}}\left( 2\sin \theta \sqrt{{{\cos }^{2}}\theta } \right) \\\ & ={{\cos }^{-1}}\left( 2\sin \theta \cos \theta \right) \\\ \end{aligned}
Now, we know that sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta ,
So, we have,
=cos⁡−1(sin⁡2θ)={{\cos }^{-1}}\left( \sin 2\theta \right)
Also, we know that,
cos⁡(π2−20)=sin⁡2θ ∴cos⁡−1(cos⁡(π2−2θ)) \begin{aligned} & \cos \left( \dfrac{\pi }{2}-20 \right)=\sin 2\theta \\\ & \therefore {{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-2\theta \right) \right) \\\ \end{aligned}
We know that cos⁡−1(cos⁡x)=x{{\cos }^{-1}}\left( \cos x \right)=x.
So, using this we have,
cos⁡−1(cos⁡(π2−2θ))=π2−2θ..........(1){{\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⁡2θx=\sin 2\theta . So, we will find the value of θ\theta from it and substitute in (1).
cos⁡−12x1−x2=π2−2θ=π2−2sin⁡−1x f(x)=π2−2sin⁡−1x \begin{aligned} & {{\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{aligned}
Now, we differentiate f(x)f\left( x \right) with the respect to x, to get the answer.
We know that,
ddxsin⁡−1(x)=11−x2\dfrac{d}{dx}{{\sin }^{-1}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}
Therefore,
ddx(π2−2sin⁡−1x) =−2ddx(sin⁡−1x) =−2(11−x2) =−21−x2 \begin{aligned} & \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{aligned}
So, the differentiation of cos⁡−12x1−x2{{\cos }^{-1}}2x\sqrt{1-{{x}^{2}}} is −21−x2\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⁡2θ=2sin⁡θcos⁡θ\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.