Apply chain rule to $u = y^{1 - n}$ in order to find $\frac{du}{dx}$ Let $u = y^{1 - n}$.
I know that, by using the chain rule:
$$\frac{du}{dx} = \frac{du}{dy} \cdot \frac{dy}{dx}$$
Also, I know that $\frac{du}{dy} = (1 - n)y^{-n} = \frac{1 - n}{y^{n}}$
Now, for $\frac{dy}{dx}$, I must re-arrange $u = y^{1 - n}$ to make $u = \frac{y}{n}$, where $nu = y$.
Hence, given that u, and n, are constant:
$$\frac{dy}{dx} = \frac{d}{dx}[un] = 0 $$
$$ \frac{du}{dx} = 0 \cdot \frac{1 - n}{y^{-n}} = 0$$
 A: You are making some inappropriate leaps. For one thing, I don't know how you went about your "rearrangement" of $u=y^{1-n},$ but you should have $$u=\frac{y}{y^n}\neq\frac{y}{n}.$$
Secondly, $u$ is explicitly a nonconstant function of $y$, and $y$ is implied to be a function of $x.$ The only way that $u$ can be constant is if $y$ is a constant function, which we must not assume.
Finally, unless you know how $y$ depends on $x,$ you cannot give a specific expression for $\frac{dy}{dx}.$ Hence, the best answer you're going to get (unless you have more context that you haven't told us about) is $$\frac{1-n}{y^n}\cdot\frac{dy}{dx}.\tag{$\star$}$$
This may seem distressing, but it might help to think of $u$ as the form that some function of $x$ might take, so our derivative of $u$ (with respect to $x$) should be just as formal. If we happen to know how $y$ depends on $x$ precisely, then the form $(\star)$ allows us to quickly calculate said derivative precisely. For example, if $u=(\ln x)^{1-n}$--that is, if $y=\ln x$--then $\frac{dy}{dx}=\frac1x,$ and so by $(\star),$ $$\frac{du}{dx}=\frac{1-n}{(\ln x)^n\cdot x}.$$ Or if $u=(\sin x)^{1-n}$--that is, if $y=\sin x$--then $\frac{dy}{dx}=\cos x,$ and so by $(\star),$ $$\frac{du}{dx}=\frac{(1-n)\cos x}{(\sin x)^n}.$$
