If $f(x)=(x-1)^2\sin{x}$, then $f'(0)=$

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To calculate the first derivative, use the product and chain rule.

$$f(x)=(x-1)^2\sin{x}$$ $$f'(x)=(x-1)^2\cos{x}+\sin{x}\cdot2(x-1)\cdot1$$ $$= (x-1)^2\cos{x}+2(x-1)\sin{x}$$

Substitute $x=0$

$$f'(0)=(0-1)^2\cos{0}+2(0-1)\sin{0}$$ $$= 1+0$$ $$=\boxed{1}$$