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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

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  • $\begingroup$ 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 $\frac{\partial f}{\partial y}$ $\endgroup$
    – Albert
    Mar 28, 2013 at 19:39
  • $\begingroup$ $y$ is independent from $x$, right? $\endgroup$
    – user27182
    Mar 28, 2013 at 19:39
  • $\begingroup$ Yes, y is independent from x $\endgroup$
    – user43758
    Mar 28, 2013 at 19:39
  • $\begingroup$ @Glougloubarbaki Ah right! $\endgroup$
    – user43758
    Mar 28, 2013 at 19:40
  • $\begingroup$ Can someone help me ? $\endgroup$
    – user43758
    Mar 28, 2013 at 19:42

2 Answers 2

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

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

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

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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.

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