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F(x) = sin(x² + 2x - 1)²

f'(x) = ?

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It depends if you mean


f(x) = \sin(x^2+2x-1)^2 = \sin^2(x^2+2x-1)

i.e. the sine part is getting squared, or


f(x) = \sin(x^2+2x-1)^2 = \sin\left((x^2+2x-1)^2\right)

i.e. the argument to sine is getting squared. I'll assume the first case, since it's fairly common convention to write
g(x)^2 = \bigg(g(x)\bigg)^2.

Now if


f(x) = \sin^2(x^2+2x-1)

• by the power and chain rules we have


f'(x) = 2 \sin(x^2 + 2x - 1) \left(\sin(x^2+2x-1)\right)'

• using the derivative of sine and the chain rule again we have


f'(x) = 2 \sin(x^2+2x-1) \cos(x^2+2x-1) \left(x^2+2x-1\right)'

• with the power and chain rules we have


f'(x) = 2 \sin(x^2+2x-1) \cos(x^2+2x-1) (2x+2)

Recalling the double angle identity for sine,


\sin(2x) = \sin(x) \cos(x)

we can rewrite the derivative among several other ways as


\boxed{f'(x) = (2x+2) \sin(2x^2+4x-2)}

User Neal Stublen
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