DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

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Inverse trigonometric functions are simply the inverse functions of trigonometric functions. They include arcsin, arccos, arctan, arcsec, arccsc, arccot. It is important to note the following.

                   Arcsin is the same as $\sin^{-1}$

                   Arccos is the same as $\cos^{-1}$

                   Arctan is the same as $\tan^{-1}$

                   Arcsec is the same as $\sec^{-1}$

                   Arcsc is the same as $\csc^{-1}$

                   Arccot is the same as $\cot^{-1}$

To differentiate inverse trigonometric functions. Here is what you need to know.

Here is what you should know

$$\frac{d}{dx}(\sin^{-1} u)=\frac{u'}{\sqrt{1-u^2}}$$

$$\frac{d}{dx}(\cos^{-1} u)=\frac{-u'}{\sqrt{1-u^2}}$$

$$\frac{d}{dx}(\tan^{-1} u)=\frac{u'}{1+u^2}$$

$$\frac{d}{dx}(\csc^{-1} u)=\frac{-us}{u\sqrt{u^2-1}}$$

$$\frac{d}{dx}(\sec^{-1} u)=\frac{u'}{u\sqrt{u^2-1}}$$

$$\frac{d}{dx}(\cot^{-1} u)=\frac{-us}{1+u^2}$$

Example 1

Differentiate $y=\sin^{-1} x +\cos^{-1} x$

Solution:

Here, u=x and u'=1

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+(\frac{-1}{\sqrt{1-x^2}})$$

$$\frac{dy}{dx}=\frac{1-1}{\sqrt{1-x^2}}=0$$

Example 2

Differentiate $y=\cot^{-1}(4x^3+3x^2)$

Solution:

Recalled that:

$$\frac{d}{dx}(\cot^{-1} u)=\frac{-us}{1+u^2}$$

Here $u=4x^3+3x^2$ and $u'=12x^2+6x$

$$\frac{dy}{dx}=-\frac{12x^2+6x}{1+(4x^3+3x^2)^2}$$

$$\frac{dy}{dx}=-\frac{12x^2+6x}{(4x^3+3x^2)^2+1}$$

If you will like to simplify further.

$$\frac{dy}{dx}=-\frac{12x^2+6x}{(4x^3+3x^2)(4x^3+3x^2)+1}$$

$$\frac{dy}{dx}=-\frac{12x^2+6x}{16x^6+24x^5+9x^4+1}$$

$$\frac{dy}{dx}=-\frac{6x(2x+1)}{x^4(4x+3)^2+1}$$

Example 3

If $h(x)=\csc^{-1}(3x)+\sec^{-1}(3x)$, Evaluate $h'(x)$

Solution:

$$h'x=\frac{-3}{3x\sqrt{(3x)^2-1}}+\frac{3}{3x\sqrt{(3x)^2-1}}$$

$$h'(x)=\frac{-3+3}{3x\sqrt{(3x)^2-1}}$$

$$h'(x)=\frac{0}{3x\sqrt{(3x)^2-1}}=0$$

Example 4

Differentiate the function $y= x\sin^{-1}x$

Solution:

$x\sin^{-1}x$ is a product, Therefore, we use the product rule which is exemplified as 

$$(fg)'=f'g+g'f$$

Here $f=x$, $f=1$, $g=\sin^{-1}x$ , $g'=\frac{1}{\sqrt{1-x^2}}$

$$\frac{dy}{dx}=1(\sin^{-1}x)+(\frac{1}{\sqrt{1-x^2}})x$$

$$\frac{dy}{dx}=1(\sin^{-1}x)+(\frac{x}{\sqrt{1-x^2}})$$

Example 5

Differentiate $y=\tan^{-1}(\sqrt{x^2+x})$

Solution:

Recalled that:

$$\frac{d}{dx}(\tan^{-1} u)=\frac{u'}{1+u^2}$$

$u=\sqrt{x^2+x}$ and $u'=\frac{2x+1}{2\sqrt{x^2+x}}$ via chain rule.

$$\frac{dy}{dx}=\frac{\frac{2x+1}{2\sqrt{x^2+x}}}{1+(\sqrt{x^2+x})^2}$$

$$\frac{dy}{dx}=\frac{\frac{2x+1}{2\qrt{x^2+x}}}{+x^2+x}$$

$$\frac{dy}{dx}=\frac{2x+1}{2\sqrt{x^2+x}}\div x^2+x+1$$

$$\frac{dy}{dx}=\frac{2x+1}{2\sqrt{x^2+x}}\times\frac{1}{x^2+x+1}$$

$$\frac{dy}{dx}=\frac{2x+1}{2\sqrt{x^2+x}(x^2+x+1)}$$

There you have it. If you have got questions relating to this post, feel free to tell me in the comment.

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Meanwhile, If you want to know more about differentiation, Read the following posts. 

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