Chain rule for 2nd derivatives

[quietrain]

hi does anyone know why the 2nd derivative chain rule is as such?

i roughly know that

if u = f(x,y) and x=rcos(T) , y = rsin(T)

then

du/dr = df/dx * dx/dr + df/dy * dy/dr

but if i am going to have a second d/dr, then how does it work out?

 

[spamiam]

I think you're mixing up the chain rule for single- and multivariable functions. For the single variable case,

<br /> (f \circ g)&#039;(x) = f&#039;(g(x))g&#039;(x)<br />

Using the product rule, we then have

<br /> (f \circ g)&#039;&#039;(x) = \frac{d}{dx}[f&#039;(g(x))g&#039;(x)] = f&#039;&#039;(g(x)) g&#039;(x) g&#039;(x) + f&#039;(g(x))g&#039;&#039;(x) = f&#039;&#039;(g(x))[g&#039;(x)]^2 + f&#039;(g(x))g&#039;&#039;(x)<br />

or in Leibniz notation,

<br /> \frac{d^2 f}{dx^2}(g(x)) \left[\frac{dg}{dx}\right]^2 + \frac{df}{dx}(g(x)) \frac{d^2 g}{dx^2}<br />

For a two-variable function things are more complicated. Suppose we have a function f(x,y) where x and y are themselves functions x(r,t) and y(r,t). As you stated,

<br /> \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r}<br />

Then

<br /> \frac{\partial^2 f}{\partial r^2} = \frac{\partial}{\partial r}\left[\frac{\partial f}{\partial x} \frac{\partial x}{\partial r}\right] + \frac{\partial}{\partial r}\left[\frac{\partial f}{\partial y} \frac{\partial y}{\partial r}\right]<br />

To make things simpler, let's just look at that first term for the moment. The tricky part is that \frac{\partial f}{\partial x} is still a function of x and y, so we need to use the chain rule again. Using the chain rule,

<br /> \frac{\partial}{\partial r}\left[\frac{\partial f}{\partial x}\right] = \frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial r} + \frac{\partial^2 f}{\partial y \partial x} \frac{\partial y}{\partial r}<br />

So

<br /> \frac{\partial}{\partial r}\left[\frac{\partial f}{\partial x} \frac{\partial x}{\partial r}\right] = \frac{\partial}{\partial r}\left[\frac{\partial f}{\partial x}\right]\frac{\partial x}{\partial r} + \frac{\partial f}{\partial x} \frac{\partial^2 x}{\partial r^2} = \left(\frac{\partial^2 f}{\partial x^2} \frac{\partial x}{\partial r} + \frac{\partial^2 f}{\partial y \partial x} \frac{\partial y}{\partial r}\right)\frac{\partial x}{\partial r} + \frac{\partial f}{\partial x} \frac{\partial^2 x}{\partial r^2}<br />

I'm feeling a bit too lazy at the moment to write out the whole thing for \frac{\partial^2 f}{\partial r^2}. (So many \frac's and \partial's...). Hopefully what I've written so far will help you enough and hopefully I didn't make any mistakes!

 

[quietrain]

ah i see... thank you very much