# How do you differentiate f(x) = 1/sqrt(arctan(2x^3)  using the chain rule?

Nov 5, 2016

$f ' \left(g \left(x\right)\right) g ' \left(x\right) = \frac{1}{2} \left(\arctan \left(2 {x}^{3}\right)\right) \cdot \frac{6 {x}^{2}}{1 + {\left(2 {x}^{3}\right)}^{2}}$

#### Explanation:

first, take the derivative of the denominator using chain rule:
$f = \sqrt{x}$
$f ' = \frac{1}{2} {x}^{- \frac{1}{2}}$
$g = \arctan \left(2 {x}^{3}\right)$
$g ' = \frac{6 {x}^{2}}{1 + {\left(2 {x}^{3}\right)}^{2}}$

$f ' \left(g \left(x\right)\right) g ' \left(x\right) = \frac{1}{2} {\left(\arctan \left(2 {x}^{3}\right)\right)}^{- \frac{1}{2}} \cdot \frac{6 {x}^{2}}{1 + {\left(2 {x}^{3}\right)}^{2}}$

that's it! if you want to simplify it further you can but my calc teacher only requires this.

note: recognize that the derivative of arctan has the chain rule built into it already if that makes sense, as $\frac{d}{\mathrm{dx}} \left[\arctan \left(u\right)\right] = \frac{u '}{1 + {u}^{2}}$