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Find the derivative of $\sin^4(x) -\cos^4(x)$.

My attempt:

$\frac{d}{dx}(\sin^4(x) -\cos^4(x)) = 4\sin^3(x)\cos(x) +4\cos^3(x)\sin(x). $

The problem is I need to simplify this to its simplest form and I just do not know how to do it to this equation.


$4\sin^3(x)\cos(x) +4\cos^3(x)\sin(x) =\sin(x)\cos(x)[4(\sin^2(x)+\cos^2(x))] = 2\sin(2x).$

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Look for a common factor.... – Ross Millikan Jan 9 '13 at 5:04
Thanks, works nicely. – Dreamer78692 Jan 9 '13 at 5:08
up vote 7 down vote accepted

A little simplification of the given expression would ease of the differentiation.

$$\sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)=-\cos 2x$$

So, $$\frac{d(\sin^4x-\cos^4x)}{dx}=\frac{d(-\cos 2x)}{dx}=+2\sin2x$$

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