A certain partial derivative

Question

Hi I am reviewing partial derivatives. For the question below, I am not sure why $(x-1)$ appears. Could anyone give me a explanation on this?

$y = x\sin(z)e^{-x}$

$\partial y/\partial x = -e^{-x}(x-1)\sin(z)$

Answer 1

Hint: Apply the product rule, then factorise the resulting expression.

Answer 2

Notice, we have $$y=x\sin (z)e^{-x}$$ $$\frac{\partial }{ \partial x}(y)=\frac{\partial}{\partial x}(x\sin (z)e^{-x})$$ $$\frac{\partial y}{\partial x}=\sin(z)\frac{\partial }{\partial x}(xe^{-x})$$ $$=\sin(z)\left(x\frac{d}{dx}(e^{-x})+\sin(z)e^{-x}\frac{d}{dx}(x)\right)$$ $$=\sin(z)\left(x(-e^{-x})+\sin(z)e^{-x}(1)\right)$$ $$=\sin(z)\left(-xe^{-x}+e^{-x}\right)$$ $$=-e^{-x}(x-1)\sin(z)$$

Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\frac{\partial y}{ \partial x}}=\color{blue}{-e^{-x}(x-1)\sin(z)}}$$

Your answer is correct.