# How do you differentiate y= 1/x (sin^-5(x)) - x/3 (cos^3(x))?

Oct 14, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {\arcsin}^{4} \left(x\right)}{x \sqrt{1 - {x}^{2}}} - {\arcsin}^{5} \frac{x}{x} ^ 2 - {\cos}^{3} \frac{x}{3} + x {\cos}^{2} \left(x\right) \sin \left(x\right)$

#### Explanation:

Assuming by ${\sin}^{- 5} \left(x\right)$ you mean ${\arcsin}^{5} \left(x\right)$ we have

$y = {\arcsin}^{5} \frac{x}{x} - \frac{x {\cos}^{3} \left(x\right)}{3}$

By saying

$y = u - v$

We can break it down into smaller derivatives and doing that such that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} - \frac{\mathrm{dv}}{\mathrm{dx}}$, so for

$u = {\arcsin}^{5} \frac{x}{x}$, we have, by the product rule that

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \cdot \frac{d}{\mathrm{dx}} \left({\arcsin}^{5} \left(x\right)\right) + {\arcsin}^{5} \left(x\right) \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \cdot \frac{d}{\mathrm{dx}} \left({\arcsin}^{5} \left(x\right)\right) - {\arcsin}^{5} \frac{x}{x} ^ 2$

By the chain rule, we have that $\frac{d}{\mathrm{dx}} \left({\arcsin}^{5} \left(x\right)\right) = 5 {\arcsin}^{4} \left(x\right) \cdot {\arcsin}^{'} \left(x\right)$ where ${\arcsin}^{'} \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{5 {\arcsin}^{4} \left(x\right)}{x \sqrt{1 - {x}^{2}}} - {\arcsin}^{5} \frac{x}{x} ^ 2$

And for

$v = \frac{x {\cos}^{3} \left(x\right)}{3}$

We have, by the product rule and the chain rule

$\frac{\mathrm{dv}}{\mathrm{dx}} = {\cos}^{3} \left(x\right) \frac{d}{\mathrm{dx}} \left(\frac{x}{3}\right) + \frac{x}{3} \frac{d}{\mathrm{dx}} \left({\cos}^{3} \left(x\right)\right)$

$\frac{\mathrm{dv}}{\mathrm{dx}} = {\cos}^{3} \frac{x}{3} + \frac{x}{3} \cdot 3 {\cos}^{2} \left(x\right) \cdot - \sin \left(x\right)$

$\frac{\mathrm{dv}}{\mathrm{dx}} = {\cos}^{3} \frac{x}{3} - x {\cos}^{2} \left(x\right) \sin \left(x\right)$

Adding it all up we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} - \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {\arcsin}^{4} \left(x\right)}{x \sqrt{1 - {x}^{2}}} - {\arcsin}^{5} \frac{x}{x} ^ 2 - {\cos}^{3} \frac{x}{3} + x {\cos}^{2} \left(x\right) \sin \left(x\right)$