# How do you differentiate sin(arctanx)?

Aug 28, 2017

$\frac{1}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{ chain rule}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\sin \left(\arctan x\right)\right)$

$= \cos \left(\arctan x\right) \times \mathrm{dx} \left(\arctan x\right)$

$= \frac{\cos \left(\arctan x\right)}{1 + {x}^{2}}$

=1/(1+x^2)xx1/(sqrt(1+x^2)

$= \frac{1}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

Aug 29, 2017

$\frac{1}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right) .$

#### Explanation:

Let, $y = \sin \left(a r c \tan x\right) = \sin \theta , \theta = a r c \tan x , \text{ so that, } \tan \theta = x .$

Now, $\tan \theta = x \Rightarrow {\csc}^{2} \theta = 1 + {\cot}^{2} \theta = 1 + \frac{1}{\tan} ^ 2 \theta ,$

$\therefore {\csc}^{2} \theta = 1 + \frac{1}{x} ^ 2 = \frac{{x}^{2} + 1}{x} ^ 2 \Rightarrow \sin \theta = \frac{1}{\csc} \theta = \frac{x}{\sqrt{{x}^{2} + 1}} .$

Hence, $y = \sin \theta = \frac{x}{\sqrt{{x}^{2} + 1}} .$

Using the Quotient Rule for Diffn.,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{{x}^{2} + 1} \frac{d}{\mathrm{dx}} \left(x\right) - x \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 1}\right)}{\sqrt{{x}^{2} + 1}} ^ 2. . . \left(\ast\right) .$

Here, by the Chain Rule,

$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 1}\right) = \frac{d}{\mathrm{dx}} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} ,$

$= \frac{1}{2} \cdot {\left({x}^{2} + 1\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) ,$

$= \frac{1}{2} \cdot {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \cdot \left(2 x\right) .$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 1}\right) = \frac{x}{\sqrt{{x}^{2} + 1}} .$

Utilising this, in $\left(\ast\right) ,$ we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{{x}^{2} + 1} \cdot 1 - x \cdot \frac{x}{\sqrt{{x}^{2} + 1}}}{{x}^{2} + 1} ,$

$= \frac{\frac{{x}^{2} + 1 - {x}^{2}}{\sqrt{{x}^{2} + 1}}}{{x}^{2} + 1} ,$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right) ,$ is the desired Diffn.