# What is the derivative of y=tan^-1 sqrt(3x)?

Jun 7, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2 \left(1 + 3 x\right) \sqrt{3 x}}$

#### Explanation:

Apply the chain rule in a straightforward manner.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {\left(\sqrt{3 x}\right)}^{2}} \times \frac{1}{2 \sqrt{3 x}} \times 3 = \frac{3}{2 \left(1 + 3 x\right) \sqrt{3 x}}$

Jun 7, 2016

$\frac{d}{\mathrm{dx}} \left(\arctan \left(\sqrt{3 x}\right)\right) = \frac{\sqrt{3}}{2 \sqrt{x} \left(3 x + 1\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\arctan \left(\sqrt{3 x}\right)\right)$

Applying chain rule,
$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
Let ,$\sqrt{3 x}$=$u$
$= \frac{d}{\mathrm{du}} \left(\arctan \left(u t\right)\right) \frac{d}{\mathrm{dx}} \left(\sqrt{3 x}\right)$
$\frac{d}{\mathrm{du}} \left(\arctan \left(u\right)\right) = \frac{1}{{u}^{2} + 1}$
$\frac{d}{\mathrm{dx}} \left(\sqrt{3 x}\right) = \frac{\sqrt{3}}{2 \sqrt{x}}$
so,
$= \frac{1}{{u}^{2} + 1} \frac{\sqrt{3}}{2 \sqrt{x}}$
substitute back,$u = \setminus \sqrt{3 x}$
$= \frac{1}{{\left(\sqrt{3 x}\right)}^{2} + 1} \frac{\sqrt{3}}{2 \sqrt{x}}$

Simplifying back,
$= \frac{\sqrt{3}}{2 \sqrt{x} \left(3 x + 1\right)}$