# What is the derivative of this function y=sec^-1(x^7)?

##### 1 Answer
Dec 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x \sqrt{{x}^{14} - 1}}$

#### Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you use the chain rule.

Let $y = {\sec}^{-} 1 \left({x}^{7}\right) \iff \sec y = {x}^{7}$

Differentiate Implicitly:

$\sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6}$
$\therefore {x}^{7} \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6}$
$\therefore \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x}$ $\left(x \ne 0\right)$ .... [1]

Using the $\sec \text{/} \tan$ identity;

${\tan}^{2} y + 1 = {\sec}^{2} y$
$\therefore {\tan}^{2} y + 1 = {\left({x}^{7}\right)}^{2}$
$\therefore {\tan}^{2} y = {x}^{14} - 1$
$\therefore \tan y = \sqrt{{x}^{14} - 1}$

Substituting into [1]
$\therefore \sqrt{{x}^{14} - 1} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x \sqrt{{x}^{14} - 1}}$