The derivative of something with respect to $3x+5$? If you take $(3x+5)^2$ and differentiate it with respect to $3x+5$ it's just $2(3x+5)$. Can someone explain to me how this would actually work out? I understand normal derivatives with respect to say, $x$, where at some point $x$, $f '(x)$ is the slope at that $x$ value. But how would this work out in this situation?
 A: It's very useful to understand that just because you have used a particular
variable $x$ in a derivative or an integral
does not mean you are stuck having to do all your derivatives
or integrals with respect to the same variable.
You can, in fact, take a derivative of the same thing but with respect
to a different variable. In both cases, you can interpret the derivative as
the slope of a curve on a graph, but which curve you use depends on what you
are differentiating over.
Here are two graphs of the quantity $(3x + 5)^2$:

In the graph on the left, $(3x+5)^2$ is plotted as a function of $x,$ much as
you might expect. In the graph on the left, the same quantity is plotted as
a function of a different variable, $u,$ which we choose to define by the
equation $u = 3x + 5.$
Because $u$ is defined that way, the graph it makes is shifted $\frac53$ units to
the right (as compared to the graph with respect to $x$) and the plot of the
function is three times as wide. (Both graphs are to exactly the same scale.)
It is sometimes very helpful to use a different variable in this way. Here we can
already see that the graph with respect to $u$ can be a little easier to work with
than the graph with respect to $x,$ since the graph over $u$ is symmetric around
the $y$-axis.
A: $\dfrac{df}{d(3x+5)}\cdot\dfrac{d(3x+5)}{dx}=\dfrac{df}{dx}$, by the chain rule. Dividing, we get:
$\dfrac{df}{d(3x+5)}=\dfrac{df}{dx}\div\dfrac{d(3x+5)}{dx}=f'(x)\div3=\dfrac{f'(x)}3$.
So, in summary, the derivative of $f$ with respect to $3x+5$ is just $\dfrac{f'(x)}3$.
A: Well, define
$$f(x)=(3x+5)^2$$
and
$$g(x)=3x+5.$$
In doing this, you are essentially calculating the value of
$$\frac{f'(x)}{g'(x)}$$
since, you are trying to know how $f(x)$ changes with a change in $g(x)$; since they both relate to $x$, you can see that by changing the value of $g(x)$ at a rate of $r$, the value $x$ will be changing at a rate of $\frac{r}{g'(x)}$ and so the value of $f(x)$ must be changing at a rate of $\frac{r}{g'(x)}f'(x)=r\frac{f'(x)}{g'(x)}$. This is an intuitive way to see it. 
Algebraically, you can always see that, if we set
$$v = f(x)$$
and
$$u = g(x)$$
then we get
$$dv = f'(x) dx$$
$$du = g'(x) dx$$
and hence
$$\frac{g'(x)}{du} = dx$$
meaning, by substitution
$$dv = \frac{f'(x)}{g'(x)} du$$
$$\frac{dv}{du} = \frac{f'(x)}{g'(x)}$$
(the fact that such manipulations work on $du$, $dx$, and $dv$ is essentially equivalent to the chain rule)
A: This is the differentiate rule of composite functions.
Here we can view $3(x+5)$ as $f(x)=u^2=(3x+5)^2$ and $u=3x+5$. Accroding the differentiate rule of composite functions, we have 
$$f'(x)=2u\times u'(x)=2(3x+5)\times 3=6(3x+5).$$
However, respect to $u$, it will be $2u$.
