Question

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

A) $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$

B) 1

C) ${{\cos }^{-1}}x$

D) ${{\tan }^{-1}}\dfrac{1}{\sqrt{1-{{x}^{2}}}}$

Answer 1

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

Complete step by step solution:

Let ${{\sin }^{-1}}x=u$ and ${{\cos }^{-1}}\sqrt{1-{{x}^{2}}}=v$

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

$\Rightarrow \dfrac{du}{dx}=\dfrac{d{{\sin }^{-1}}x}{dx}$

We know the differentiation of ${{\sin }^{-1}}x$ is equal to $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$

Therefore,

$\Rightarrow \dfrac{du}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$…………….. $\left( 1 \right)$

Now, we will differentiate v with respect to x.

$\Rightarrow \dfrac{dv}{dx}=\dfrac{d{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}}{dx}$

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

Therefore, the equation becomes.

$\Rightarrow \dfrac{dv}{dx}=-\dfrac{\dfrac{-2x}{2\sqrt{1-{{x}^{2}}}}}{\sqrt{1-{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}}$

Simplifying it further, we get

$\Rightarrow \dfrac{dv}{dx}=-\dfrac{\dfrac{-2x}{2\sqrt{1-{{x}^{2}}}}}{\sqrt{{{x}^{2}}}}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$…………….. $\left( 2 \right)$

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

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

Therefore,

$\Rightarrow \dfrac{du}{dv}=\dfrac{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}$

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

$\Rightarrow \dfrac{du}{dv}=1$

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

Therefore, the derivative of ${{\sin }^{-1}}x$ with respect to ${{\cos }^{-1}}\sqrt{1-{{x}^{2}}}$ 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 $f(g(x))$ is equal to $f'(g(x)).g'(x)$