What is $\frac {d}{dx}(y=\frac {e^{-x/2}}{u^{1/2}})$? I'm not sure if I need to use the chain rule here or not. I saw a video on YouTube where someone found that the $\frac {dy}{dx}$ of $y=xz$ is:
$$\frac {dy}{dx} = x\frac {dz}{dx} + z$$
So I feel like I am more on track with the second one. Could someone explain how to take this derivative?
EDIT - I'm being asked if $u(x)$, but I'm not sure. The original problem is a differential equation that I need to solve using substitution:
$$
ydx + (1+ye^x)dy = 0
$$
It was suggested that I use the substitution $u=e^{-x}/y^2$, so, after solving for $y$, I need to find its derivative. Hence this question. Does this help clarify things? Is $u(x)$?
 A: What the yotube video has done is assumed that $z(x)$, so it has a derivative with respect to $x$. I will answer your question one step at a time, assuming that $u(x)$.
$$\frac{d}{dx} \left (y=\frac{e^{-\frac{x}{2}}}{u^{\frac{1}{2}}}\right)$$
Per comment, move $u$ to numerator to apply multiplication rule:
$$\frac{d}{dx} \left (y=e^{-\frac{x}{2}}*u^{-\frac{1}{2}}\right)$$
$$\frac{dy}{dx} = \frac{d}{dx} \left (e^{-\frac{x}{2}}*u^{-\frac{1}{2}}\right)$$
Multiplication rule says:$\frac{d}{dx}(AB)=\frac{dA}{dx}B+A\frac{dB}{dx}$. So apply it to your problem:
$$\frac{dy}{dx} = \frac{d}{dx}(e^{-\frac{x}{2}})*u^{-\frac{1}{2}}+e^{-\frac{x}{2}}*\frac{d}{dx} (u^{-\frac{1}{2}})$$
Use exponential rule for first term: $\frac{d}{dx} e^{u(x)}=e^{u(x)}*\frac{d}{dx}u(x)$. So apply it to our problem:
$$\frac{dy}{dx} = -\frac{1}{2}(e^{-\frac{x}{2}})*u^{-\frac{1}{2}}+e^{-\frac{x}{2}}*\frac{d}{dx} (u^{-\frac{1}{2}})$$
Apply power rule to our problem: $\frac{d}{dx} u(x)^a=a*u(x)^{a-1}*\frac{d}{dx}u(x)$. So apply it to our problem:
$$\frac{dy}{dx} = -\frac{1}{2}(e^{-\frac{x}{2}})*u^{-\frac{1}{2}}+e^{-\frac{x}{2}}*-\frac{1}{2}(u^{-\frac{3}{2}})*\frac{du}{dx}$$
