# What is the derivative of this function arcsec(x^3)?

Nov 16, 2017

$= \frac{3 x}{\sqrt{{x}^{2} - 1}}$

#### Explanation:

For this you need to know what the derivative of arcsec(x) is.

You can derive it fairly easily:

$y = a r c \sec \left(x\right)$
$\sec \left(y\right) = x$
$\frac{d}{\mathrm{dx}} \sec \left(y\right) = \frac{d}{\mathrm{dx}} x$
$\sec \left(y\right) \tan \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
dy/dx = 1/(sec(y)tan(y)
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec \left(a r c \sec \left(x\right)\right) \tan \left(a r c \sec \left(x\right)\right)}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x \sqrt{{x}^{2} - 1}}$

In the last statement I simplified $\sec \left(a r c \sec \left(x\right)\right)$ to $x$, which is obvious, but I also simplified $\tan \left(a r c \sec \left(x\right)\right)$ to $\sqrt{{x}^{2} - 1}$ which is slightly less obvious

If you draw a triangle such that sec(angle) = x, you can understand why it works:

you can see that $a r c \sec \left(x\right) = \theta$ and $\tan \left(\theta\right) = \sqrt{{x}^{2} - 1}$

$\therefore \tan \left(a r c \sec \left(x\right)\right) = \sqrt{{x}^{2} - 1}$

Now we know the derivative of arcsec(x) and just need to apply a little chainrule to get our answer:

$\frac{d}{\mathrm{dx}} a r c \sec \left({x}^{3}\right) = \frac{1}{x \sqrt{{x}^{2} - 1}} \cdot \frac{d}{\mathrm{dx}} {x}^{3} = \frac{3 {x}^{\cancel{2}}}{\cancel{x} \sqrt{{x}^{2} - 1}}$
$= \frac{3 x}{\sqrt{{x}^{2} - 1}}$

Nov 16, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\left\mid x \right\mid \sqrt{{x}^{6} - 1}}$

#### Explanation:

If $y = a r c \sec \left({x}^{3}\right)$ then $\sec y = {x}^{3}$, that is: $\cos y = \frac{1}{x} ^ 3$

Differentiate implicitly:

$- \sin y \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{x} ^ 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{{x}^{4} \sin y}$

Now, for $0 \le y \le \pi$:

$\sin y = \sqrt{1 - {\cos}^{2} y} = \sqrt{1 - \frac{1}{x} ^ 6}$

and:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{{x}^{4} \sqrt{1 - \frac{1}{x} ^ 6}} = \frac{3 \sqrt{{x}^{6}}}{{x}^{4} \sqrt{{x}^{6} - 1}} = \frac{3}{\left\mid x \right\mid \sqrt{{x}^{6} - 1}}$