Partial derivative of $f(x,y)=g(2x+5y)$ Let $g:\mathbb{R}\to\mathbb{R}$ be a differentiable function of one variable and let
$f(x,y)=g(2x+5y)$.
How do I find the partial derivative $f_x(x,y)$? I am more interested in an explanation than the result. Which rule (chain rule?) should I use etc., and how?
I know that the answer has form $f_x(x,y)=\text{[some number]}g'(2x+5y)$.
Update:
I now get:
$\frac{\partial g(2x+5y)}{\partial x}=\frac{\partial g(x,y)}{\partial (2x+5y)}\frac{\partial (2x+5y)}{\partial x}=\frac{\partial g(2x+5y)}{\partial (2x+5y)}2=g'(2x+5y)2$
Is this correct?
 A: example:
$f(x)=x^2. $
$f'(x) = 2x.$ 
$Now, f(2x+5y) = (2x+5y)^2. $
$then, f'(2x+5y) = 2(2x+5y) * \frac{\partial}{\partial x}(2x+5y) = 2(2x+5y) * 2. $
This is sort of like if I had the original $f'(x)$, then replaced $x$ with $2x+5y$, then multiplied by $2$.
That's due to the chain rule.  You'll notice, since $g(x)$ is a composite function with $2x+5y$, whatever $g(x)$'s derivative originally was, I can get the derivative of the composite function by sticking in $2x+5y$ and then multiplying by the derivative of that (which is $2$).
A: I find differentials easier to calculate with since, unlike partial derivatives, the notation says what it means, and their calculation fairly directly mirrors how you would compute derivatives in single variable calculus; in particular, recall how implicit differentiation works. So I would first compute
$$\mathrm{d} f(x,y)
= \mathrm{d} g(2x + 5y)
= g'(2x + 5y) \mathrm{d}(2x + 5y)
= g'(2x + 5y) (2 \mathrm{d}x + 5 \mathrm{d}y) $$
In terms of differentials, the intent of the notation $f_x(x,y)$ is to refer to the result you get if you compute $\mathrm{d}f(x,y)$ and substitute $\mathrm{d}x \to 1$ and $\mathrm{d}y \to 0$. Thus,
$$ f_x(x,y) = g'(2x + 5y) \cdot 2$$
