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How do you find the derivative of #sin^3(2x) / x#?

Answer 1

Evelyn Agard

In this way, using the division rule:

#y'=(3sin^2 2x * cos2x * 2 *x-sin^3 2x*1)/x^2=#

#=(sin^2 2x(6xcos2x-sin2x))/x^2#.

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Answer 2

Evelyn Agard

To find the derivative of (\frac{{\sin^3(2x)}}{x}), use the quotient rule:

Let (u = \sin^3(2x)) and (v = x).

Apply the quotient rule: [ \frac{{d}}{{dx}}\left(\frac{{\sin^3(2x)}}{x}\right) = \frac{{v \frac{{du}}{{dx}} - u \frac{{dv}}{{dx}}}}{{v^2}}]

Now, find (\frac{{du}}{{dx}}) and (\frac{{dv}}{{dx}}): [ \frac{{du}}{{dx}} = 3\sin^2(2x)\cdot 2\cos(2x)\cdot 2 = 12\sin^2(2x)\cos(2x)] [ \frac{{dv}}{{dx}} = 1]

Substitute these derivatives into the quotient rule: [ \frac{{d}}{{dx}}\left(\frac{{\sin^3(2x)}}{x}\right) = \frac{{x\cdot 12\sin^2(2x)\cos(2x) - \sin^3(2x)\cdot 1}}{{x^2}}]

Simplify: [ \frac{{d}}{{dx}}\left(\frac{{\sin^3(2x)}}{x}\right) = \frac{{12x\sin^2(2x)\cos(2x) - \sin^3(2x)}}{{x^2}}]

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Answer from HIX Tutor

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