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Question 28

Number 28

The first derivative of $y = \cos^3 x$ is ....

$3\cos x \cdot \sin x$

$-3\cos x \cdot \sin x$

$2\cos x \cdot \sin 2x$

$-6\cos x \cdot \sin 2x$

$\frac{-3}{2}\cos x \cdot \sin 2x$

Explanation

Given $y = \cos^3 x$

Derivative Using Chain Rule

Use the chain rule to differentiate the composite function

y' = 3(\cos^2 x) \cdot (-\sin x)

= -3\cos^2 x \sin x

= -3\cos x \cos x \sin x

Using Trigonometric Identity

Use the identity $2\sin x \cos x = \sin 2x$

= \frac{-3}{2}\cos x (2\sin x \cos x)

= \frac{-3}{2}\cos x \cdot \sin 2x

Therefore, the first derivative of $y = \cos^3 x$ is $y' = \frac{-3}{2}\cos x \cdot \sin 2x$

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Question 28Set 1

Exercises

Number 28

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