Question

What is the derivative of ${\sin ^{ - 1}}(x)$?

Answer 1

Here, in the question given, we are asked to find the derivative of inverse of sine function. We will first put the given value equal to some unknown variable $y$. This will help us in differentiating the given value with respect to its variable $x$. And then we will further simplify it to get the desired result. We will use the trigonometric identities, if needed.
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step by step solution:
Let $y = {\sin ^{ - 1}}x$
Taking $\sin$ both sides, we get,
$\sin y = \sin \left( {{{\sin }^{ - 1}}x} \right)$
Simplifying the right hand side, we get,
$\sin y = x\;\;\; \ldots \left( 1 \right)$ where $- \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$ (according to sine inverse definition)
Now, differentiating w.r.t. $x$ both sides, we get
$\dfrac{d}{{dx}}\left( {\sin y} \right) = \dfrac{d}{{dx}}\left( x \right)$
Simplifying it, we get
$\cos y\dfrac{{dy}}{{dx}} = 1$
Dividing by $\cos y$ both sides, we obtain,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos y}}$
Since $y \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, we have positive values of $\cos y$
Using the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\sin }^2}y} }}$
Substituting the value of $y$ from equation $\left( 1 \right)$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Substituting the value of $y$ back again, we obtain
$\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}$

Note: The symbol ${\sin ^{ - 1}}x$ should not be confused with ${\left( {\sin x} \right)^{ - 1}}$. In-fact ${\sin ^{ - 1}}x$ is an angle, the value of whose sine is $x$. The only key concept to solve such types of questions is that we must remember all the basic trigonometric identities and their application. This will help us to solve almost all the questions.
The restriction on $y$ taken above is there only to be sure that we get a consistent answer out of inverse of sine. We know that there are infinite numbers of angles that will work but we want to get the consistent value when we work with the inverse of the sine function.