# Derivatives of Inverse trig functions- Free Mathematics Tips

In this section we will see the derivatives of the inverse trigonometric functions. To derive the derivatives of inverse trigonometric functions we will need the previous formala’s of derivatives of inverse functions

#### Formula to find derivatives of inverse trig function.

If $$f\left(x\right)$$ and $$g\left(x\right)$$are inverse trig functions then,

$$g’\left(x\right)=\frac{1}{f’\left(g\left(x\right)\right)}$$

We Know that two functions are inverses if ,$$f\left(g\left(x\right)\right)=x$$ and $$g\left(f\left(x\right)\right)=x$$

### Derivatives of inverse trig functions- Sine

Let’s start with inverse sine. Here is the definition of the inverse sine.

$$\color{black} {y=\sin^{-1}\Leftrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for \ \ \ \ \ \ \ -\frac{\pi}{2}\leq y\leq\frac{\pi}{2}}$$

Evaluate:- $$\displaystyle {\sin ^{ – 1}}\left( {\frac{1}{2}} \right)$$

we know the relation between inverse sine and sine function ;

$$\sin \left( {{{\sin }^{ – 1}}x} \right) = x\hspace{0.5in}{\sin ^{ – 1}}\left( {\sin x} \right) = x$$

They are inverses for each other. So we can use this to find out the derivative of inverse sine function

$$f\left( x \right) = \sin x\hspace{0.5in}g\left( x \right) = {\sin ^{ – 1}}x$$

Then, $$g’\left( x \right) = \frac{1}{{f’\left( {g\left( x \right)} \right)}} = \frac{1}{{\cos \left( {{{\sin }^{ – 1}}x} \right)}}$$,

This is not a better formula . Let’s start by the definition of the inverse sine function.

$$y = {\sin ^{ – 1}}\left( x \right)\hspace{0.5in} \Rightarrow \hspace{0.5in}x = \sin \left( y \right)$$,

$$\cos \left( {{{\sin }^{ – 1}}x} \right) = \cos \left( y \right)$$,

$${\cos ^2}y + {\sin ^2}y = 1\hspace{0.5in} \Rightarrow \hspace{0.5in}\cos y = \sqrt {1 – {{\sin }^2}y}$$,

$$\cos \left( {{{\sin }^{ – 1}}x} \right) = \cos \left( y \right) = \sqrt {1 – {{\sin }^2}y}$$,

$$\cos \left( {{{\sin }^{ – 1}}x} \right) = \cos \left( y \right) = \sqrt {1 – {{\sin }^2}y}$$,

$$\frac{d}{{dx}}\left( {{{\sin }^{ – 1}}x} \right) = \frac{1}{{\sqrt {1 – {x^2}} }}$$ Derivatives of inverse trig functions

Here is a proof of derivative of cot x

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