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Question: The derivative of \({{\sin }^{-1}}x\) w.r.t. \({{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\) is \(\left( -1\le ...

The derivative of sin⁡−1x{{\sin }^{-1}}x w.r.t. cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}} is (−1≤x≤1)\left( -1\le x\le 1 \right)

A) 11−x2\dfrac{1}{\sqrt{1-{{x}^{2}}}}

B) 1

C) cos⁡−1x{{\cos }^{-1}}x

D) tan⁡−111−x2{{\tan }^{-1}}\dfrac{1}{\sqrt{1-{{x}^{2}}}}

Explanation

Solution

We have to find the derivative of sin⁡−1x{{\sin }^{-1}}x with respect to cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}. For that, we will first find the differentiation of sin⁡−1x{{\sin }^{-1}}x with respect to x and then we will find the differentiation of cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}} with respect to x. Then we will divide the value of the derivative of cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}} obtained by the value of derivative of sin⁡−1x{{\sin }^{-1}}x. From there, we will get the result of the derivative of sin⁡−1x{{\sin }^{-1}}x with respect to cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}.

Complete step by step solution:

Let sin⁡−1x=u{{\sin }^{-1}}x=u and cos⁡−11−x2=v{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}=v

Now, we will first differentiate u with respect to x.

⇒dudx=dsin⁡−1xdx\Rightarrow \dfrac{du}{dx}=\dfrac{d{{\sin }^{-1}}x}{dx}

We know the differentiation of sin⁡−1x{{\sin }^{-1}}x is equal to 11−x2\dfrac{1}{\sqrt{1-{{x}^{2}}}}

Therefore,

⇒dudx=11−x2\Rightarrow \dfrac{du}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}…………….. (1)\left( 1 \right)

Now, we will differentiate v with respect to x.

⇒dvdx=dcos⁡−11−x2dx\Rightarrow \dfrac{dv}{dx}=\dfrac{d{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}}{dx}

We know the differentiation of cos⁡−1x{{\cos }^{-1}}x is equal to−11−x2-\dfrac{1}{\sqrt{1-{{x}^{2}}}}, so to find differentiation of cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}, we will use chain rule of differentiation.

Therefore, the equation becomes.

⇒dvdx=−−2x21−x21−(1−x2)2\Rightarrow \dfrac{dv}{dx}=-\dfrac{\dfrac{-2x}{2\sqrt{1-{{x}^{2}}}}}{\sqrt{1-{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}}

Simplifying it further, we get

⇒dvdx=−−2x21−x2x2=11−x2\Rightarrow \dfrac{dv}{dx}=-\dfrac{\dfrac{-2x}{2\sqrt{1-{{x}^{2}}}}}{\sqrt{{{x}^{2}}}}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}…………….. (2)\left( 2 \right)

According to the question, we have to find the derivative of sin⁡−1x{{\sin }^{-1}}x with respect to cos⁡−11−x2 {{\cos }^{-1}}\sqrt{1-{{x}^{2}}}

For that, we will divide equation (1)\left( 1 \right) by equation (2)\left( 2 \right)

Therefore,

⇒dudv=11−x211−x2\Rightarrow \dfrac{du}{dv}=\dfrac{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}

As the value of numerator and denominator are the same. So its division will be equal to one.

⇒dudv=1\Rightarrow \dfrac{du}{dv}=1

We have found dudv\dfrac{du}{dv} which is equal to derivative of sin⁡−1x{{\sin }^{-1}}x with respect to cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}

Therefore, the derivative of sin⁡−1x{{\sin }^{-1}}x with respect to cos⁡−11−x2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}} is 1.

Hence, option (B) is correct.

Note:

We need to know the following terms as we have used them in this solution.

Differentiation is defined as a process in which we find the function which gives the output of rate of change of one variable with respect to another variable.

Differentiation by chain rule: In this method, the differentiation of function f(g(x))f(g(x)) is equal to f′(g(x)).g′(x)f'(g(x)).g'(x)