# How do you differentiate  g(x) = sin (arctan (x/sqrt(3))) ?

Jan 9, 2018

$\frac{d}{\mathrm{dx}} \left(\sin \left(\arctan \left(\frac{x}{\sqrt{3}}\right)\right)\right) = \frac{3}{{x}^{2} + 3} ^ \left(\frac{3}{2}\right)$

#### Explanation:

Let's first simplify $\sin \left(\arctan \left(\frac{x}{\sqrt{3}}\right)\right)$.

We know that ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$. Divide both sides by $\sin {\left(\theta\right)}^{2}$ to get $1 + \frac{1}{\tan} ^ 2 \left(\theta\right) = \frac{1}{\sin} ^ 2 \left(\theta\right)$, or ${\sin}^{2} \left(\theta\right) = \frac{1}{1 + \frac{1}{\tan} ^ 2 \left(\theta\right)}$.

Thus, $\sin \left(\theta\right) = \sqrt{\frac{1}{1 + \frac{1}{\tan} ^ 2 \left(\theta\right)}}$.

So, using the above identity, $\sin \left(\arctan \left(\frac{x}{\sqrt{3}}\right)\right) = \sqrt{\frac{1}{1 + \frac{1}{\tan} ^ 2 \left(\arctan \left(\frac{x}{\sqrt{3}}\right)\right)}}$.

Simplify:
$= \sqrt{\frac{1}{1 + \frac{1}{{x}^{2} / 3}}} = \frac{x}{\sqrt{{x}^{2} + 3}}$.

To differentiate this, use the quotient rule $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{\frac{\mathrm{du}}{\mathrm{dx}} v - \frac{\mathrm{dv}}{\mathrm{dx}} u}{v} ^ 2$ to find the final answer

$\frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 3}}\right) = \frac{\sqrt{{x}^{2} + 3} - {x}^{2} / \sqrt{{x}^{2} + 3}}{{x}^{2} + 3} = \frac{3}{{x}^{2} + 3} ^ \left(\frac{3}{2}\right)$