find $f^{\text{'}}(x)f(x)=sin(-x^2+1)$

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

$f^{\text{'}}(x)=f^{\text{'}}(g(x))*g^{\text{'}}(x)$

stip $1)f^{\text{'}}(g(x))=cos*1*(-x^2+1{)}^{1-1}=cos$

stepz$).g^{\text{'}}(x)=-x^2+1=-1*2*x^{2-1}=-2x$

Step $3).f^{\text{'}}(x)=cos(x)*-2x$ why are we including when $(-x^2+1{)}^{1-1}=0$

step $4)f^{\text{'}}(x)=-2xcos(x)$

Real answer is: $-2xcos(-x^2+1)?$

$>>>>>$the answer my prof. Got is differnet from what i got and im confused why, i need a answer thats simple and not complex please :)