# Differentiate sin^-1 x

The inverse trigonometric function are represented by adding arc in prefix for a trigonometric function, or by adding the power of -1, such as:

Inverse of sin x = arcsin(x) or $\sin^{-1}x$

Given $y = \sin^{-1}x$…………(i)

$\Rightarrow x = \sin y$

Differentiating the above equation w.r.t. x, we have:

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\cos y}$

Putting the value of y form (i), we get

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\cos y} = \frac{1}{\cos (\sin^{-1}x)}$………..(ii)

From equation (ii), we can see that the value of cos y cannot be equal to 0, as the function would become undefined.

$\Rightarrow \sin^{-1}x \neq \frac{-\pi}{2}, \frac{\pi}{2}$ i.e. $x \neq -1,1$

From (i) we have

$y = \sin^{-1}x$ $\Rightarrow \sin y = \sin (\sin^{-1}x)$

Using property of trigonometric function,

$\cos^{2}y = 1 – \sin^{2}y = 1 – (\sin (\sin^{-1}x))^{2} = 1 – x^{2}$ $\Rightarrow \cos y = \sqrt{1 – x^{2}}$…………(iii)

Now putting the value of (iii) in (ii), we have

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\sqrt{1-x^{2}}}$

Therefore, the Derivative of Inverse sine function is

$\frac{\mathrm{d} }{\mathrm{d} x} (\sin^{-1}x)= \frac{1}{\sqrt{1-x^{2}}}$