Calculating partial derivatives Let f and g be functions of one real variable and define $F(x,y)=f[x+g(y)]$. Find formulas for all the partial derivatives of F of first and second order.
For the first order, I think we have:
$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}$
$\frac{\partial F}{\partial y}=\frac{\partial f}{\partial x}g'(x)+ \frac{\partial f}{\partial y}g'(y)$
Is it correct? What are the second order derivatives?
Thank you
 A: $f$ is a function of one variable. Therefore the notation $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ is problematic (and I suggest you adapt the prime notation in that case). What you have written is not correct.
The correct formulas are: $$\frac{\partial F}{\partial x}(x,y)=f'(x+g(y)) $$
$$\frac{\partial F}{\partial y}(x,y)=f'(x+g(y))g'(y)  $$ $$\frac{\partial^2 F}{\partial x^2}(x,y)=f''(x+g(y)) $$ $$\frac{\partial^2 F}{\partial x \partial y}(x,y)=f''(x+g(y))g'(y)=\frac{\partial^2 F}{\partial y \partial x}(x,y) $$
$$\frac{\partial^2 F}{\partial y^2}(x,y)=f''(x+g(y))g'(y)+f'(x+g(y))g''(y) $$
A: you need to differentiate $f$ by its argument, then differentiate the argument by $x$ or $y$
Setting $\xi = x + g(y)$,
$\frac{d F}{d x}=\frac{df}{d \xi} \frac{d \xi}{dx} = \frac{df}{d \xi}$
and 
$\frac{d F}{d y}=\frac{df}{d \xi} \frac{d \xi}{dy} = \frac{df}{d \xi} g'(y)$
You seem to have looked up the chain rule, but just didn't notice that $f$ has one argument only, so you can probably do the $2^{nd}$ order ones ok.
