Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $\frac{dy}{dx}=18$ when $x=2$, find $f'(4)$ 
Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $\frac{dy}{dx}=18$ when $x=2$, find $f'(4)$.

I'm sorry if I'm asking this beginner question, but I'm really confused about how to solve this problem. I've tried to substitute $u=4$ and other things that I thought might work. I'm kinda stuck.
I've got $f'(4)=-\frac{27}6$ but in the textbook it is $-\frac9{40}$.
 A: Differentiate with respect to $x$:
$$y'=2(f(u)+3x)(f'(u)u'+3)=2(f(x^3-2x)+3x)(f'(x^3-2x)(3x^2-2)+3)$$
Substitute the known values at $x=2$ and simplify:
$$18=2(f(4)+6)(10f'(4)+3)=2\cdot12(10f'(4)+3)$$
$$\frac{18}{24}=10f'(4)+3$$
$$f'(4)=\frac{3/4-3}{10}=-\frac9{40}$$
So the textbook is right after all.
A: You have $u(2)=4$ so $f(u(2))=f(4)=6$. Let's write $\frac{dy}{dx}$:
$$\begin{align}\frac{dy}{dx}&=2(f(u)+3x)\frac d{dx}(f(u)+3x)\\
&=2(f(u)+3x)\left(\frac{df(u(x))}{dx}+3\right)\\&=2(f(u)+3x)\left(\frac{df(u)}{du}\frac{du}{dx}+3\right)\\&=2(f(u)+3x)\left(f'(u)(3x^2-2)+3\right)\end{align}$$
Now just plug in $\frac{dy}{dx}=18$, $u=4$, $f(4)=6$, and $x=2$.
A: Note that for $x=2, u=4$
$$\begin{align}y&=(f(u(x))+3x)^2\\\dfrac{\mathrm d}{\mathrm dx}(y)&=\dfrac{\mathrm d}{\mathrm dx}(f(u(x))+3x)^2\\\dfrac{\mathrm dy}{\mathrm dx}&=2(f(u(x))+3x)\cdot\dfrac{\mathrm d}{\mathrm dx}(f(u(x))+3x)\\&=2(f(u(x))+3x)(f'(u(x))\cdot u'(x)+3)\\\text{Setting }&x=2...\\18&=2(f(u(2))+6)(f'(u(2))\cdot u'(2)+3)\\9&=(f(4)+6)(f'(4)\cdot(3\cdot2^2-2)+3)\\9&=12(10f'(4)+3)\\\dfrac34&=10f'(4)+3\\10f'(4)&=-\dfrac94\\f'(4)&=-\dfrac9{40}\end{align}$$
