How do you take the derivative of a variable with respect to a different variable: How can I use the chain rule $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ to differentiate an expression like $\frac{d}{dx}(y^3)$ ?   The solution in the book is 
\begin{equation*}
\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}.  
\end{equation*}
I'm not sure how they got it though.
 A: In the problem, you want to differentiate the expression $\color{green}y^{\color{blue}3}$ with the variable $\color{red}x$. But there is no information on what $\color{green}y$ is. So we must assume $\color{green}y$ is an arbitrary function, i.e. $\color{green}y := \color{green}y(\color{red}x)$. 
Now the crucial point is to note that the 3 itself is also a function, i.e. we can define $\color{blue}z(\color{green}y) = \color{green}y^{\color{blue}3}$.
In this way, we could apply the chain rule (some names changed to be consistent with the problem),
$$ 
\frac{d \color{blue} z}{d \color{red} x} = \frac{d \color{blue} z}{d \color{green} y} \cdot \frac{d \color{green} y}{d \color{red} x}
$$
i.e. explictly,
$$
\frac{d \Bigl( \color{blue} z\bigl(\color{green}y(\color{red}x)\bigr) \Bigr)}{d \color{red} x}
= \frac{d \Bigl( \color{blue} z \bigl(\color{green}y\color{silver}{(x)}\bigr) \Bigr) }{d\bigl( \color{green}y\color{silver}{(x)}\bigr)}
\cdot \frac{d \bigl( \color{green} y (\color{red}x)\bigr) }{d \color{red} x}
$$
In $d\color{blue}z/d\color{green}y$, since no other variables (like $\color{red}x$) is directly involved, we could treat $\color{green}y$ as a variable instead of an implicit function. In $d\color{green}y/d\color{red}x$, as the function $\color{green}y(\color{red}x)$ is arbitrary, the differentiated form is the best representation already. Therefore, the expression above simplifies to
$$
\frac{d \left( \color{green}y^{\color{blue}3} \right) }{d \color{red} x}
= \underbrace{  \color{teal}{ 3 \color{green}y^2 }  }_{d(\color{green}y^{\color{blue}3})/d\color{green}y} \cdot 
\frac{d \color{green} y}{d \color{red} x}
$$
A: I think there is a bit of a mix up over notation - specifically, the problem is using $y$ in two different contexts.
The definition of the chain rule that you need (and what you have correctly stated):
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
Now, for a composite function $y(u(x)) = [u(x)]^3$ (here, $y$ is the 'cubing' function), in order to take the derivative of $y$ with respect to $x$ we first take the derivative of $y$ w.r.t. $u$, then since $u$ is a function of $x$ (and should be written as $u(x)$ at this point) we use the chain rule as follows:
$\frac{dy}{du} = 3u^{2}$, and $\frac{du}{dx} = u'(x)$ since we don't have the expression of the function $u(x)$ itself. So it looks like
$$\frac{d}{dx}([u(x)]^3) = 3[u(x)]^{2}u'(x).$$
This looks different because I have chosen to relabel the function $y^3$ as $u^{3}$ in order for the stated chain rule to 'look right.'  Alternatively, we could state the chain rule as $\frac{dh}{dx} = \frac{dh}{dy}\frac{dy}{dx}$ where $h = h(y(x)) = [y(x)]^3$.  Then $\frac{dh}{dy} = 3y^2$ and $\frac{dy}{dx}$ is just $y'$ again because we don't have a specific expression for that function.  Here we would have $\frac{dh}{dx} = 3y^2y'$.
