Second Partial Differentiation - Explain this to me please!

**gva0324** · November 18th 2009, 10:41 AM

I need help with the explanation for this one.

If T (r)

and $r = \sqrt{x^2+ y^2}$

This makes

$\frac{\partial T}{\partial x} = \frac{x}{r}\frac{dT}{dr}$

Now I need to find the second derivative:

Using operators

$\frac{\partial}{\partial x}(T) = \frac{x}{r}\frac{d}{dr}(T)$

For the second derivative it would become

$\frac{\partial}{\partial x}(\frac{dT}{dx})$ = $\frac{\partial}{\partial x}(\frac{x}{r}\frac{dT}{dr})$

I understand upto this point, but then I do not understand how this turns up becoming:

$\frac{x}{r}\frac{\partial}{\partial r}(\frac{x}{r}\frac{dT}{dr})$

Can somebody please tell me what happened in between??

Thank you very much!!

**Danny** · November 18th 2009, 10:56 AM

Originally Posted by gva0324

I need help with the explanation for this one.

If T (r)

and $r = \sqrt{x^2+ y^2}$

This makes

$\frac{\partial T}{\partial x} = \frac{x}{r}\frac{dT}{dr}$

Now I need to find the second derivative:

Using operators

$\frac{\partial}{\partial x}(T) = \frac{x}{r}\frac{d}{dr}(T)$

For the second derivative it would become

$\frac{\partial}{\partial x}(\frac{dT}{dx})$ = $\frac{\partial}{\partial x}(\frac{x}{r}\frac{dT}{dr})$

I understand upto this point, but then I do not understand how this turns up becoming:

$\frac{x}{r}\frac{\partial}{\partial r}(\frac{x}{r}\frac{dT}{dr})$

Can somebody please tell me what happened in between??

Thank you very much!!

It might be easier to see the general chain rule

$<br /> T_x = T_r \cdot r_x<br />$

$<br /> T_{xx} = T_{rr} \cdot \left(r_x\right)^2 + T_r \cdot r_{xx}.<br />$

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