1. ## Advanced Proof of Chain Rule

assuming differentiability of functionsthe substitution t=g(x,y) converts F(t) into f(x,y) where f(x,y)=F[(g(x,y))]
Show that

$\displaystyle \frac{\partial f}{\partial x}=F'[(g(x,y))]\frac{\partial g}{\partial x}, \frac{\partial f}{\partial y}=F'[(g(x,y))]\frac{\partial g}{\partial y}$

am I supposed to use a first order Taylor function with error term?
for instance h(a+y)-h(a)=f[g(a+y)]-f[g(a)]=f(b+v)-f(b) can be used to prove that h is differentiable at a (I guess it means that the limit exists) and the total derivative h'(a) is equal to the composition f'(b)=g'(a)

2. ## Re: Advanced Proof of Chain Rule

I think you need to go back to the definition of the derivative (as a limit of the difference quotient). For the first equation, start with:

$\displaystyle \frac{f(x,y_0)-f(x_0,y_0)}{x-x_0} = \frac{F(g(x,y_0))-F(g(x_0,y_0))}{g(x,y_0)-g(x_0,y_0)}\cdot\frac{g(x,y_0)-g(x_0,y_0)}{x-x_0}$

The left-hand side and the second fraction on the right-hand side are good. There's a problem with the first fraction on the right-hand side, though, since $\displaystyle g(x,y_0)$ could be equal to $\displaystyle g(x_0,y_0)$ for any or even all x in an interval around $\displaystyle x_0$. The trick is to define a function Q(y) which is equal to $\displaystyle \frac{F(y)-F(g(x_0,y_0))}{y-g(x_0,y_0)}$ when $\displaystyle y \ne g(x_0,y_0)$ and $\displaystyle f'(g(x_0,y_0))$ when $\displaystyle y=g(x_0,y_0)$. So you have:

$\displaystyle \frac{f(x,y_0)-f(x_0,y_0)}{x-x_0} = Q(g(x,y_0))\cdot\frac{g(x,y_0)-g(x_0,y_0)}{x-x_0}$

because when $\displaystyle g(x,y_0) \ne g(x_0,y_0)$, the $\displaystyle g(x,y_0)-g(x_0,y_0)$ terms cancel, and when $\displaystyle g(x,y_0) = g(x_0,y_0)$, both sides are zero. Now you can take the limit as x goes to $\displaystyle x_0$.

- Hollywood

3. ## Re: Advanced Proof of Chain Rule

Thanks so much! this is p.275 in vol. 2 Apostol, I'm not sure if I should try analysis instead. My lungs are failing and don't work so well.... so I guess it's adequate to say that when $\displaystyle y=g(x_0,y_0)$, both sides are 0 since you multiply the left by the denominator of the right, therefore the limit is 0. And when $\displaystyle g(x,y_0) \ne g(x_0,y_0)$ then you have holding y as constant
$\displaystyle \lim_{x\rightarrow x_0}\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0}=\frac{\partial f}{\partial x}= \lim_{x\rightarrow x_0} \frac{F(g(x,y_0))-F(g(x_0,y_0))}{x-x_0}=F'[g(x,y)]\frac{\partial g}{\partial x}$

and also holding x constant we can define a function Q(x) similarly as well so that

$\displaystyle \lim_{y\rightarrow y_0}\frac{f(x_0,y_)-f(x_0,y_0)}{y-y_0}=\frac{\partial f}{\partial y}= \lim_{y\rightarrow y_0} \frac{F(g(x_0,y))-F(g(x_0,y_0))}{y-y_0}=F'[g(x,y)]\frac{\partial g}{\partial y}$

and for the special case where
$\displaystyle F(t)=e^{sint}, g(x,y)=cos(x^{2}+y^{2}), \frac{\partial f}{\partial x}=-e^{sin(cos(x^{2}+y^{2}))}cos(cos(x^{2}+y^{2}))sin( x^{2}+y^{2}))(2x)$
which has level curves of concentric circles about the origin