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The process of finding the derivatives of a function by using the differentiation from first principles or by delta method. In this section we will find the derivatives of

$sin^{-1}(x), cos^{-1}(x),tan^{-1}(x),sec^{-1}(x),csc^{-1}(x) and cot^{-1}(x)$.

Derivatives of inverse trigonometric functions

**1) If f(x) = $sin^{-1}(x) = arcsin(x)$ $x\in (-1,1) and \frac{-\pi}{2} \leq f(x)\leq \frac{\pi}{2}$ then,
f '(x) = $\frac{1}{\sqrt{1-x^{2}}}$
2) If f(x) = $cos^{-1}(x) = arccos(x)$ $x\in (-1,1) and 0 \leq f(x) \leq \pi$ then,
f '(x) = $\frac{-1}{\sqrt{1-x^{2}}}$
3) If f(x) = $tan^{-1}(x) = arctan(x),\frac{-\pi}{2} < f(x) < \frac{\pi}{2}$ then,
f '(x) = $\frac{1}{1+x^{2}}$
4) If f(x) = $cot^{-1}(x) = arccot(x),0 < f(x) < \pi$ then,
f '(x) = $\frac{-1}{1+x^{2}}$
5) If f(x) = $sec^{-1}(x) = arcsec(x) $
$x\in R -[-1,1] , 0 \leq f(x) \leq \pi and f(x)\neq \frac{\pi}{2} $ then,
f '(x) = $\frac{1}{x\sqrt{x^{2}-1}}$
6) If f(x) = $csc^{-1}(x) = arccsc(x)$ $x\in R -[-1,1] , \frac{-\pi}{2} \leq f(x) \leq \frac{\pi}{2} and
f(x) \neq 0$ then,
f '(x) = $\frac{-1}{x\sqrt{x^{2}-1}}$**

** Prove that : If $x \in (-1,1)$,then the differentiation of $sin^{-1}x$ with respect to x is $\frac{1}{\sqrt{1-x^{2}}}$
i.e $\frac{\text{d}}{\text{d}x}(sin^{-1}x) = \frac{1}{\sqrt{1-x^{2}}}$
Proof : ** Let y = $sin^{-1}x$

Take sine on both side so we get,

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

Sin(y) = x

Now differentiate with respect to y,

$\frac{\text{d}}{\text{d}y}(sin(y)) = \frac{\text{d}x}{\text{d}y}$

Cos(y) = $\frac{\text{d}x}{\text{d}y}$ -----(1)

As we know that $sin^{2}\theta + cos^{2}\theta$ = 1

So $sin^{2}y + cos^{2}y$ = 1

$cos^{2}y = 1 - sin^{2}y $

$cos^{2}y = 1 - x^{2}$

$ cos(y) = \sqrt{1 - x^{2}}$

So equation (1) becomes,

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

$\frac{\text{d}y}{\text{d}x} = \frac{1}{\frac{\text{d}x}{\text{d}y}}$

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

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

** Prove that : If $x \in (-1,1)$,then the differentiation of $cos^{-1}x$ with respect to x is $-\frac{1}{\sqrt{1-x^{2}}}$
i.e $\frac{\text{d}}{\text{d}x}(cos^{-1}x) = -\frac{1}{\sqrt{1-x^{2}}}$
Proof : ** Let y = $cos^{-1}x$

x = cos(y)

Now differentiate with respect to x and use a chain rule ,

1 = $\frac{\text{d}}{\text{d}x}(cos(y))$

1= -sin(y). $\frac{\text{d}y}{\text{d}x} $

$\frac{\text{d}y}{\text{d}x} = \frac{1}{-sin(y)}$ ---(1)

As we know that $sin^{2}\theta + cos^{2}\theta$ = 1

So $sin^{2}y + cos^{2}y$ = 1

$sin^{2}y = 1 - cos^{2}y $

$sin^{2}y = 1 - x^{2}$

$ sin(y) = \sqrt{1 - x^{2}}$

$ sin(y) = \sqrt{1 - x^{2}}$

So equation (1) becomes,

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

$\frac{\text{d}}{\text{d}x}(cos^{-1}x) = \bf-\frac{1}{\sqrt{1-x^{2}}}$

Here we will apply the chain rule as well as derivative of inverse trigonometric function.

$\frac{\text{d}}{\text{d}x}(arcsin(2x))=\frac{1}{\sqrt{1-(2x)^{2}}}.\frac{\text{d}}{\text{d}x}(2x)$

= $\frac{1}{\sqrt{1-4x^{2}}}.2 $

$\frac{\text{d}}{\text{d}x}(arcsin(2x))=\frac{\text{d}}{\text{d}x}sin^{-1}(2x)=\frac{2}{\sqrt{1-4x^{2}}}$

** Example 2 :** f(x) = arccos(x-1)

**Solution :** f(x) = arccos(x-1)

Here we will apply the chain rule as well as derivative of inverse trigonometric function.

$\frac{\text{d}}{\text{d}x}(arccos(x-1))=\frac{\bf-1}{\sqrt{1-(x-1)^{2}}}.\frac{\text{d}}{\text{d}x}(x-1)$

= $\frac{\bf-1}{\sqrt{1-(x-1)^{2}}}.1 $

$\frac{\text{d}}{\text{d}x}(arccos(x-1))=\frac{\text{d}}{\text{d}x}cos^{-1}(x-1)=\frac{\bf-1}{\sqrt{1-(x-1)^{2}}}$

** Example 3 :** f(x) = $x^{2}.arctan(x)$

**Solution :** f(x) = $x^{2}.arctan(x)$

Here we will apply the product rule as well as derivative of inverse trigonometric function.

$\frac{\text{d}}{\text{d}x}(x^{2}.arctan(x))=arctan(x).\frac{\text{d}}{\text{d}x}x^{2} + x^{2}.\frac{\text{d}}{\text{d}x}(arctan(x))$

= $arctan(x).2x + x^{2}.\frac{1}{1 + x^{2}}$

$\frac{\text{d}}{\text{d}x}(x^{2}.arctan(x))= 2x.arctan(x) + \frac{x^{2}}{1 + x^{2}}$

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