2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ sorry, it's not correct. since $f$ is a function of one variable, it makes no sense to write $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ $\endgroup$ – Glougloubarbaki Mar 28 '13 at 19:39
  • $\begingroup$ $y$ is independent from $x$, right? $\endgroup$ – user27182 Mar 28 '13 at 19:39
  • $\begingroup$ Yes, y is independent from x $\endgroup$ – Carpediem Mar 28 '13 at 19:39
  • $\begingroup$ @Glougloubarbaki Ah right! $\endgroup$ – Carpediem Mar 28 '13 at 19:40
  • $\begingroup$ Can someone help me ? $\endgroup$ – Carpediem Mar 28 '13 at 19:42
5
$\begingroup$

$f$ 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: $$\frac{\partial F}{\partial x}(x,y)=f'(x+g(y)) $$

$$\frac{\partial F}{\partial y}(x,y)=f'(x+g(y))g'(y) $$ $$\frac{\partial^2 F}{\partial x^2}(x,y)=f''(x+g(y)) $$ $$\frac{\partial^2 F}{\partial x \partial y}(x,y)=f''(x+g(y))g'(y)=\frac{\partial^2 F}{\partial y \partial x}(x,y) $$

$$\frac{\partial^2 F}{\partial y^2}(x,y)=f''(x+g(y))g'(y)+f'(x+g(y))g''(y) $$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.