Question

Prove that the derivative of ${\sin ^{ - 1}}x$ is $\dfrac{1}{{\sqrt {1 - {x^2}} }}$.

Answer 1

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

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

Now, we know that the derivative of $\sin \theta $ with respect to $\theta $ is $\cos \theta $. So, we get,
\[ \Rightarrow \dfrac{{dx}}{{d\theta }} = \cos \theta \]
So, substituting the value of $\dfrac{{d\left( \theta \right)}}{{dx}}$ in equation $\left( 1 \right)$, we get,
$ \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 $\dfrac{1}{{\cos \theta }}$. But we need to convert it back into the given variable x.
So, we have, $\sin \theta = x$.

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