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?
 A: \begin{align}
\partial_u f(1,2) &= \lim_{h\to 0} \frac{f((1,2) + hu) - f(1,2)}h \\
&= \lim_{h\to 0} \frac{f\left(1+\frac{h}2,2+\frac{h\sqrt{3}}2\right) - f(1,2)}h\\
&= \lim_{h\to 0} \frac{g\left(15 + h\left(\frac32 + 3\sqrt3\right)\right) - g(15)}h\\
&= \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)\\
&= g'(15)\left(\frac32 + 3\sqrt3\right)\\
&= 3+6\sqrt3
\end{align}
A: By the chain rule $$\frac{\partial f}{\partial x}(x,y)=\frac{dg}{dt}(3x+6y)\frac{\partial(3x+6y)}{\partial x}$$
and $$\frac{\partial f}{\partial y}(x,y)=\frac{dg}{dt}(3x+6y)\frac{\partial(3x+6y)}{\partial y}$$
Hence $$\nabla f(1,2)=\Big(3g'(15),6g'(15) \Big)=(6,12)$$
The partial derivatives above are calculated with respect to the standard basis of $\Bbb R^2$. The partial derivative along the direction $u=(\frac12,\frac{\sqrt{3}}{2})$ is given by 
$$\frac{\partial f}{\partial u}(1,2)=\langle \nabla f(1,2), u\rangle=3+6\sqrt{3}$$
A: Directional derivative of $f$ in direction $u$:


*

*$\frac{\partial{f}}{\partial u} = grad f \cdot u$


Gradient of $f(x,y)= g(z)$ with $z = 3x+6y$:


*

*$grad f = g'(z)(\frac{\partial{z}}{\partial x}\;\; \frac{\partial{z}}{\partial y}) = g'(z)(3\;\; 6)$


You have $z(1,2)= 15$. So, you get


*

*$grad f(1,2) = g'(15)(3\;\; 6) = 2(3\;\; 6) = (6\;\; 12)$


Finally,
$$\frac{\partial{f(1,2)}}{\partial u}= (6 \;\; 12)\begin{pmatrix} \frac{1}{2}  \\ \frac{\sqrt{3}}{2} \end{pmatrix} = 3+6\sqrt{3}$$
A: By chain rule [$\frac {\partial f} {\partial x }=g'(3x+6y)\frac {\partial (3x+6y)} {\partial x}$] we get  $\frac {\partial f} {\partial x } (1,2)=3g'(15)$ because $3x+6y=15$ when $x=1$ and $y=2$. Similarly $\frac {\partial f} {\partial y } (1,2)=6g'(15)$. I suppose you can take it from here. 
A: A lot of the common notation for partial derivatives just gets in the way, I find. Let’s untangle things a bit by introducing the function $h: (x,y)\mapsto 3x+6y$ so that $f=g\circ h$. We then have by the chain rule $$\nabla f(x,y) = g'(h(x,y)) \, \nabla h(x,y).$$ We have $h(1,2)=15$ and we’re given that $g'(15)=2$, while $\nabla h$ is the constant vector $(3,6)$. Plugging this into the above formula, we get $$\nabla f(1,2) = 2(3,6)=(6,12).$$ I assume that you can take it from here.
