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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.

Answer and Explanation:

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}


Learn more about this topic:

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When to Use the Quotient Rule for Differentiation

from Math 104: Calculus

Chapter 8 / Lesson 8
51K

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