Question

F(x) = sin(x² + 2x - 1)²

f'(x) = ?

Need Best Answer? Give the explanation!
ok thx.
​

Answer 1

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)}`