Practice: differentiation

Calculus Level 1

Let f ( x ) = cos ⁡ 2 x + sin ⁡ 2 x f(x)=\cos^2x+\sin^2x , find f ′ ( x ) f'(x) .

4 cos ⁡ x + 2 cos ⁡ 2 x sin ⁡ 2 x + 4 sin ⁡ x 4\cos x+2\cos^2x\sin^2x+4\sin x 1 1 0 0 cos ⁡ 2 ( sin ⁡ 2 x ) \cos^2(\sin^2x)

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4 solutions

It's really simple if you assume you were given the identity cos ⁡ 2 x + sin ⁡ 2 x = 1 \cos^2x+\sin^2x=1 . Then the derivative of a constant is 0 0 and thus f ′ ( x ) = 0 f'(x)=0 .

If not, you may pursue like this: d d x ( cos ⁡ 2 x + sin ⁡ 2 x ) = d d x ( cos ⁡ x cos ⁡ x ) + d d x ( sin ⁡ x sin ⁡ x ) = cos ⁡ x ( − sin ⁡ x ) + ( − sin ⁡ x ) cos ⁡ x + cos ⁡ x sin ⁡ x + cos ⁡ x sin ⁡ x by the product rule = 2 cos ⁡ x sin ⁡ x − 2 cos ⁡ x sin ⁡ x = 0. \begin{aligned} \dfrac{\mathrm d}{\mathrm dx}(\cos^2x+\sin^2x)&=\dfrac{\mathrm d}{\mathrm dx}(\cos x\cos x)+\dfrac{\mathrm d}{\mathrm dx}(\sin x\sin x) \\ &=\cos x(-\sin x)+(-\sin x)\cos x+\cos x\sin x+\cos x\sin x \quad\text{by the product rule}\\ &=2\cos x\sin x -2\cos x\sin x \\ &=0. \end{aligned}

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Please, note that although perfectly correct, doing anything besides understanding that cos^2(x) + sin^2(x) is a constant is a waste of time and should be avoided at all costs during a test or exam.

Bernardo Sulzbach - 6 years, 12 months ago

If you want to actually manually find the derivative, just use the chain rule, don't even do the product rule...

Hobart Pao - 5 years, 7 months ago
Oli Hohman
Mar 23, 2015

f'(x) = -2 cos(x) sin(x) + 2*sin(x)cos(x)

[cos(x) sin(x)] (-2+2) = 0

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Sunny Sahu
Jul 12, 2014

­ I f y o u w e r e g i v e n t h e a s s u m p t i o n t h a t s i n 2 x + c o s 2 x = 1 , t h e d e r i v a t i v e o f f ( x ) w o u l d b e e a s y t o a p p r o a c h f o r t h e u s e r ( s ) f ( x ) = 1 , d d x ( 1 ) = 0 If\quad you\quad were\quad given\quad the\quad assumption\quad that\quad sin^{ 2 }x\quad +\quad cos^{ 2 }x=1,\quad the\quad derivative\quad of\quad f(x)\quad would\quad be\quad easy\quad to\quad approach\quad for\quad the\quad user(s)\\ f(x)=1,\quad \frac { d }{ dx } (1)=0

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cos^2 x +sin^2 x =1 .Thus, d/dx(1) =0

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