# How do you differentiate y=sec^-1(x^7)?

Mar 23, 2017

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

#### Explanation:

$y = {\sec}^{- 1} {x}^{7}$

$\implies {x}^{7} = \sec y$

differentiate wrt $x$

$7 {x}^{6} = \sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}}$

(dy)/(dx)=(7x^6)/(secytany

we now substitute back to express the derivative as a function of $x$

$\sec y = {x}^{7}$

$1 + {\tan}^{2} y = {\sec}^{2} y$

tany=sqrt(sec^2y-1

(dy)/(dx)=(7x^6)/(x^7sqrt(sec^2y-1)

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

Mar 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x \sqrt{{x}^{14} - 1}}$, or $\frac{7}{{x}^{8} \sqrt{1 - \frac{1}{x} ^ 14}}$

#### Explanation:

Let

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

Then:

$\sec y = {x}^{7}$

Differentiating Implicitly:

$\sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6}$
$\therefore {x}^{7} \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6}$
$\therefore \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6} / {x}^{7}$
$\therefore \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x}$

And using the identity ${\sec}^{2} \theta \equiv 1 + {\tan}^{2} \theta$ then;

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

Substituting we get:

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

Which can also be written:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{x \sqrt{{x}^{14} \left(1 - \frac{1}{x} ^ 14\right)}}$
$\text{ } = \frac{7}{x \cdot {x}^{7} \cdot \sqrt{1 - \frac{1}{x} ^ 14}}$
$\text{ } = \frac{7}{{x}^{8} \sqrt{1 - \frac{1}{x} ^ 14}}$