Chain rule for 2nd derivatives

  • Thread starter quietrain
  • Start date
  • #1
654
2
hi does anyone know why the 2nd derivative chain rule is as such?

a1c3a6d5dafe24a2f4c2d423a8e87fdb.png


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?
 

Answers and Replies

  • #2
360
0
I think you're mixing up the chain rule for single- and multivariable functions. For the single variable case,

[tex]
(f \circ g)'(x) = f'(g(x))g'(x)
[/tex]

Using the product rule, we then have

[tex]
(f \circ g)''(x) = \frac{d}{dx}[f'(g(x))g'(x)] = f''(g(x)) g'(x) g'(x) + f'(g(x))g''(x) = f''(g(x))[g'(x)]^2 + f'(g(x))g''(x)
[/tex]

or in Leibniz notation,

[tex]
\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}
[/tex]

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,

[tex]
\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}
[/tex]

Then

[tex]
\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]
[/tex]

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

[tex]
\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}
[/tex]

So

[tex]
\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}
[/tex]

I'm feeling a bit too lazy at the moment to write out the whole thing for [itex] \frac{\partial^2 f}{\partial r^2} [/itex]. (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!
 
  • #3
654
2
ah i see... thank you very much
 

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