chain rule misunderstanding In my assignment, I am asked to evaluate the derivative of $4x(3x+5)^3$, using the chain rule. 
I believe that there are three levels the this equation: $4x$, $x^3$, and $3x+5$, which I will call f(x), g(x) and h(x). 
So, I find that $f'(g(h(x))) = 4$ , $g'(h(x))=3(x)^2)$, and $h'(x) = 3)$. 
I then multiply these together: $4(3x+5)^3 * 3(3x+5)^2 * 3$
However, the answer in my book is $4(3x + 5)^2(12x + 5)$
I'm just learning to use the chain rule, where does my misunderstanding lie? 
 A: If $f(x) = 4x$, $g(x) = x^3$, and $h(x) = 3x+5$, then $f(g(h(x))) = 4(3x+5)^3$ not $4x(3x+5)^3$.
You probably want to use the product rule.
A: You must use the product rule, as stated in the other answer. Specifically, use the fact that
$\dfrac d{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x)$
where $f(x) = 4x$ and $g(x) = (3x + 5)^3$.
But let me give you some additional insight on the chain rule that some teachers overlook.
We can think about $\dfrac {dy}{dx}$ as $\dfrac {dy}{dx} * 1 = \dfrac {dy}{dx} * \dfrac {du}{du} = \dfrac {dy}{du} * \dfrac {du}{dx} $, where $du$ is just some arbitrary differential term derived from some arbitrary expression involving some variable $u$.
In this case we can imagine $y = (3x + 5)^3$ as $y = u^3$ where $u = 3x + 5$. Now, we can find $\dfrac {du}{dx}$ easily; it is just $3$. But we are not done; we still need $\dfrac {dy}{du}$. This is also fairly easily; it's just $3u^2$. So therefore, $\dfrac {dy}{dx} = \dfrac {dy}{du} \dfrac {du}{dx} = 3u^2 * 3 = 9u^2$. But $u$ is just some variable we invented; we want our answer in terms of $x$. But alas! We defined $u = 3x + 5$, so we can substitute that in for $u$, for a result of $\dfrac {dy}{dx} = 9(3x+5)^2$.
A: When you use Chain Rule, you should always check to make sure you've chosen the functions correctly. In your case, you set $f(x)=4x,g(x)=x^3,$ and $h(x)=3x+5$. You would want your expression to then be $f\big(g(h(x))\big)$, but plugging in we get 
$$f\big(g(h(x))\big)=4(3x+5)^3$$
You're close, but remember that in function composition, you aren't multiplying, you're substituting.
In order to solve this, we'll need more than chain rule. Notice we're actually multiplying the two outer parts, $4x$ and $(3x+5)^3$, so we'll the product rule. Using this, we get
$$\frac{d}{dx}f\big(g(h(x))\big) = \frac{d}{dx}4x * (3x+5)^3 + 4x*\frac{d}{dx}(3x+5)^2$$
That left part we can do easy enough, that's just $4(3x+5)^3$ The right is where we'll need Chain Rule. By setting $f(x)=x^3$ (the outer function) and $g(x)=3x+5$ (the inner function), we have $f(g(x))=(3x+5)^3$, like we want. Now just like before:
$$
\begin{align}
f'(g(x))&=3(3x+5)^2\\
g'(x)&=3\\
\frac{d}{dx}f(g(x))&=g'(x)f'(g(x))\\
&=3*3(3x+5)^2\\
&=9(3x+5)^2
\end{align}
$$
Plugging this back into what we have before from product rule, and we get
$$
\begin{align}
\frac{d}{dx}4x(3x+5)^3&=4(3x+5)^3+4x*9(3x+5)^2\\
&=4(3x+5)^3+36x(3x+5)^2\\
&=(3x+5)^2(4(3x+5)+36x)\\
&=(3x+5)^2(12x+20+36x)\\
&=(3x+5)^2(48x+20)
\end{align}
$$
