 # Derivative of Inverse Trigonometric functions

The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. These functions are used to obtain angle for a given trigonometric value. Inverse trigonometric functions have various application in engineering, geometry, navigation etc.

## Representation of functions:

Generally, 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 $$\sin^{-1}x$$

Let us now find the derivative of Inverse trigonometric function

Example: Find the derivative of a function $$y = \sin^{-1}x$$.

Solution: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

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

## Derivatives of Inverse trigonometric functions

 Function ($$\frac{\mathrm{d} y}{\mathrm{d} x}$$) arcsin x $$\frac{1}{\sqrt{1-x^{2}}}$$ arccos x $$-\frac{1}{\sqrt{1-x^{2}}}$$ arctan x $$\frac{1}{1+ x^{2}}$$ arccot x $$-\frac{1}{1+ x^{2}}$$ arcsec x $$\frac{1}{\left | x \right |\sqrt{x^{2}-1}}$$ arccsc x $$-\frac{1}{\left | x \right |\sqrt{x^{2}-1}}$$
 Example:Find the derivative of a function 2 arcsin x – 5 arccsc x. Solution:Given $$2\; arcsin x – 5\; arccsc x$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2}{\sqrt{1-x^{2}}} + \frac{5}{ x \sqrt{x^{2}-1}}$$ Further we can factorize the given expression. Example:Find the derivative of a function $$\sin^{-1} \left ( \frac{1-x^{2}}{1+x^{2}} \right )$$. Solution:Given y = $$\sin^{-1} \left ( \frac{1-x^{2}}{1+x^{2}} \right )$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\sqrt{1-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}} \times \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{1-x^{2}}{1+x^{2}} \right )$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\sqrt{ \frac{(1+x^{2})^{2}- (1-x^{2})^{2}}{(1+x^{2})^{2}}} } \times \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{1-x^{2}}{1+x^{2}} \right )$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1+x^{2}}{\sqrt{ (1+x^{4}+2x^{2})- (1+x^{4}-2x^{2})} } \times \left ( \frac{(1+x^{2})(-2x)- (1-x^{2})(2x))}{(1+x^{2})^{2}} \right )$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1+x^{2}}{\sqrt{ 4x^{2}} } \times \left ( \frac{(-2x -2x^{3}- 2x + 2x^{3}) }{(1+x^{2})^{2}} \right )$$ $$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1+x^{2}}{2x} \times \left ( \frac{-4x }{(1+x^{2})^{2}} \right )$$ $$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-2}{1+x^{2}}$$
Quiz on Derivative of Inverse Trigonometric functions

#### 1 Comment

1. Nice explain 😃 thank you so much for all byjus Teachers 🙏🖤