# What's the derivative of f(x) = sin (arctan (x/(sqrt(3))))?

Apr 1, 2016

$\cos \left(\arctan \left(\frac{x}{\sqrt{3}}\right)\right) \cdot \frac{1}{\sqrt{3}} \cdot \frac{1}{1 + {x}^{2} / 3}$

#### Explanation:

By the chain rule:

$y = \sin u , u = \arctan \left(\frac{x}{\sqrt{3}}\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = \cos u , \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\sqrt{3}} \cdot \frac{1}{1 + {x}^{2} / 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= \cos \left(\arctan \left(\frac{x}{\sqrt{3}}\right)\right) \cdot \frac{1}{\sqrt{3}} \cdot \frac{1}{1 + {x}^{2} / 3}$

Apr 1, 2016

$f ' \left(x\right) = \frac{3}{\left({x}^{2} + 3\right) \sqrt{{x}^{2} + 3}}$

#### Explanation:

With $\theta = \arctan \left(\frac{x}{\sqrt{3}}\right)$ we get sin theta = x/(sqrt(x^2+3).

I can't easily draw a picture of the triangle here, but there is an algebraic way to get this.
$\tan \theta = \frac{x}{\sqrt{3}}$ and $- \frac{\pi}{2} < \theta < \frac{\pi}{2}$
$\Rightarrow$ ${\sec}^{2} \theta = {\tan}^{2} \theta + 1 = {x}^{2} / 3 + 1 = \frac{{x}^{2} + 3}{3}$
$\Rightarrow$ ${\cos}^{2} \theta = \frac{3}{{x}^{2} + 3}$
$\Rightarrow$ ${\sin}^{2} \theta = 1 - {\cos}^{2} \theta = 1 - \frac{3}{{x}^{2} + 3} = {x}^{2} / \left({x}^{2} + 3\right)$
$\Rightarrow$ sin theta = x/(sqrt(x^2+3).

So, we have

$f \left(x\right) = \frac{x}{\sqrt{{x}^{2} + 3}}$.

Use the quotient, power and chain rules to get

$f ' \left(x\right) = \frac{3}{\left({x}^{2} + 3\right) \sqrt{{x}^{2} + 3}}$