Directional derivative of function with two variables containing a function and its derivative. Given $f(x,y)=g(3x+6y)$ and suppose $g'(15)=2$, then I would like to take the directional derivative at the point $(1,2)$ and in the direction of $u=\left(\frac12,\frac{\sqrt{3}}{2}\right)$. 

Normally this would not be a problem but I can not see through what is going on in the function $f$. How am I supposed to interpret it?
\begin{align}
\partial_u f(1,2) &amp;= \lim_{h\to 0} \frac{f((1,2) + hu) - f(1,2)}h \\
&amp;= \lim_{h\to 0} \frac{f\left(1+\frac{h}2,2+\frac{h\sqrt{3}}2\right) - f(1,2)}h\\
&amp;= \lim_{h\to 0} \frac{g\left(15 + h\left(\frac32 + 3\sqrt3\right)\right) - g(15)}h\\
&amp;= \lim_{h\to 0} \frac{g\left(15 + h\left(\frac32 + 3\sqrt3\right)\right) - g(15)}{h\left(\frac32 + 3\sqrt3\right)} \cdot\left(\frac32 + 3\sqrt3\right)\\
&amp;= g'(15)\left(\frac32 + 3\sqrt3\right)\\
&amp;= 3+6\sqrt3
\end{align}
I now realise, that you guys were referring to the standard chain rule for one variable while I was trying to make sense of it all with the chain rule for multiple variables. All I had to do was consider the function like so: $f(g(3x+6y),g(3x+6)$. And then i could apply $'(f(g(x))=f'(g(x)*g'(x)$.