# Differentiate the given functions: y= \frac {\sin^3x}{x} \ \text {and} \ z= \frac {\cos^3x}{x}

## Question:

Differentiate the given functions: {eq}y= \frac {\sin^3x}{x} \ \text {and} \ z= \frac {\cos^3x}{x} {/eq}

## Differentiation using the quotient rule

This example illustrates the process of differentiating a function using the quotient rule.

Specifically, the quotient rule will be used alongside the chain rule, and also with standard results for the differentiation of trigonometric functions.

For the first function, we will use a combination of the quotient rule, which says that {eq}\dfrac{d}{dx}\left ( \dfrac{u}{v} \right ) = \dfrac{u'v - uv'}{v^2} {/eq}, and the chain rule, and the standard result {eq}\dfrac{d}{dx}(\sin x) = \cos x. {/eq}

So we have:

{eq}\dfrac{dy}{dx} = \dfrac{3(\sin x)^2(\cos x)\times x - \sin^3x \times 1}{x^3} = \dfrac{3x\sin^2x\cos x - \sin^3x}{x^3} {/eq}

And for the second function, we will use the quotient rule (as stated above), and the chain rule, and the standard result {eq}\dfrac{d}{dx}(\cos x) = -\sin x. {/eq}

So we have:

{eq}\dfrac{dz}{dx} = \dfrac{3(\cos x)^2(-\sin x)\times x - \cos^3x \times 1}{x^3} = \dfrac{-3x\cos^2x\sin x - \cos^3x}{x^3} {/eq}