Power Rule Isn't Sufficient

Calculus Level 1

Which of the following identities is correct?

d d x cos 2 x = cos 2 x \frac d{dx} \cos^2 x = \cos 2x d d x sin 2 x = sin 2 x \frac d{dx} \sin^2 x = \sin 2x

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4 solutions

Prince Loomba
Sep 25, 2016

Relevant wiki: Differentiation - Chain Rule

Recall that by chain rule , we have d d x ( f ( x ) ) n = n f ( x ) ( f ( x ) ) n 1 \dfrac{d}{dx} (f(x))^n = n \cdot f'(x) \cdot (f(x))^{n-1} . In this case n = 2 n=2 and f ( x ) = sin x f(x) = \sin x or f ( x ) = cos x f(x) = \cos x .

And note that d d x sin x = cos x \dfrac d{dx} \sin x = \cos x and d d x cos x = sin x \dfrac d{dx} \cos x = -\sin x . See derivatives of trigonometric functions .

d d x sin 2 x = 2 sin x d d x sin x = 2 sin x cos x = sin 2 x \frac {d}{dx}\sin^{2}x=2\sin x \frac {d}{dx}\sin x=2\sin x \cos x=\sin2x

While

d d x cos 2 x = 2 cos x d d x cos x = 2 cos x ( sin x ) = sin 2 x \frac {d}{dx}\cos^{2}x=2\cos x \frac {d}{dx}\cos x=2\cos x (-\sin x)=-\sin2x

So option 2 2 is right

Galen Buhain
Sep 28, 2016

d/dx sin^2 (x) = 2 sin (x) cos (x) = sin (2x)

Joe Potillor
Nov 9, 2016

Colorful solution. +1

Pi Han Goh - 4 years, 7 months ago

d/dx sin^2 (x) = 2 sin (x) cos (x) = sin (2x)

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