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Question: Prove that the derivative of \({\sin ^{ - 1}}x\) is \(\dfrac{1}{{\sqrt {1 - {x^2}} }}\)....

Prove that the derivative of sin⁡−1x{\sin ^{ - 1}}x is 11−x2\dfrac{1}{{\sqrt {1 - {x^2}} }}.

Explanation

Solution

In the given problem, we are required to differentiate sin⁡−1x{\sin ^{ - 1}}x with respect to x and get its value equal to 11−x2\dfrac{1}{{\sqrt {1 - {x^2}} }}. Since, sin⁡−1x{\sin ^{ - 1}}x is an inverse trigonometric function, it is quite difficult to differentiate. So, we will assume the function sin⁡−1x{\sin ^{ - 1}}x as a variable and apply the chain rule of differentiation. So, differentiation of sin⁡−1x{\sin ^{ - 1}}x with respect to x will be done layer by layer using the chain rule of differentiation. Also the derivative of sin⁡(x)\sin (x) with respect to xx must be remembered.

Complete step by step answer:
To find the derivative of sin⁡−1x{\sin ^{ - 1}}x with respect to xx we have to find the difference between sin⁡−1x{\sin ^{ - 1}}xwith respect to xx.
So, Derivative of sin⁡−1x{\sin ^{ - 1}}x with respect to xxcan be calculated as ddx(sin⁡−1x)\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) .
Now, ddx(sin⁡−1x)\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)
Substituting x as sin⁡θ\sin \theta in the expression, we get,
⇒ddx(sin⁡−1x)=ddx(sin⁡−1(sin⁡θ))\Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {\sin \theta } \right)} \right)
Now, we know that sin⁡−1(sin⁡θ){\sin ^{ - 1}}\left( {\sin \theta } \right) is equal to θ\theta itself. Hence, we get,
⇒ddx(sin⁡−1x)=d(θ)dx−−−−(1)\Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{{d\left( \theta \right)}}{{dx}} - - - - \left( 1 \right)
Now, we know that sin⁡θ=x\sin \theta = x. So, differentiating both sides of this assumption made by us with respect to θ\theta , we get,
⇒dxdθ=d(sin⁡θ)dθ\Rightarrow \dfrac{{dx}}{{d\theta }} = \dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }}

Now, we know that the derivative of sin⁡θ\sin \theta with respect to θ\theta is cos⁡θ\cos \theta . So, we get,
⇒dxdθ=cos⁡θ\Rightarrow \dfrac{{dx}}{{d\theta }} = \cos \theta
So, substituting the value of d(θ)dx\dfrac{{d\left( \theta \right)}}{{dx}} in equation (1)\left( 1 \right), we get,
⇒ddx(sin⁡−1x)=1cos⁡θ−−−−(2)\Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\cos \theta }} - - - - \left( 2 \right)
So, we get the value of the derivative as 1cos⁡θ\dfrac{1}{{\cos \theta }}. But we need to convert it back into the given variable x.
So, we have, sin⁡θ=x\sin \theta = x.

We know the trigonometric identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1. So, substituting the value of sin⁡θ\sin \theta in the identity to find the value of cosine, we get,
⇒x2+cos⁡2θ=1\Rightarrow {x^2} + {\cos ^2}\theta = 1
⇒cos⁡2θ=1−x2\Rightarrow {\cos ^2}\theta = 1 - {x^2}
Taking square root on both sides of equation, we get,
⇒cos⁡θ=1−x2\Rightarrow \cos \theta = \sqrt {1 - {x^2}}
Now, substituting the value of cosine in equation 22 to get to the required answer,
∴ddx(sin⁡−1x)=11−x2\therefore \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}
So, the derivative of sin⁡−1x{\sin ^{ - 1}}x is 11−x2\dfrac{1}{{\sqrt {1 - {x^2}} }}.
Hence, proved.

Note: The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. We should always remember the substitution that we make so as to use it in the later part of the solution.