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Chain Rule with Partial Derivatives Question

I'm having trouble getting from u_s to u_(ss). Could someone elaborate on that part a great deal. I'm pretty sure I get the rest. It's just the partial differentiation algebra which confuses me.

Even for the u_s part: I'm pretty sure f_x is a partial derivative but is x_s also a apartial derivative (to use the first summand of u_s as an example)?

Any input would be greatly appreciated!

Thanks in advance!

Re: Chain Rule with Partial Derivatives Question

The important thing to remember is that both $\displaystyle f'_x(x,y)$ and $\displaystyle f'_y(x,y)$ depend on both x and y, and in turn s and t.

$\displaystyle u=f(x,y)=f(e^s\cos t, e^s \sin t)$

$\displaystyle u'_s = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=f'_xe^s \cos t + f'_y e^s\sin t$

$\displaystyle u''_{ss}=\frac{\partial }{\partial s}\left (f'_xe^s \cos t + f'_y e^s\sin t \right) = \frac{\partial }{\partial s}\left( f'_xe^s \cos t \right)+\frac{\partial }{\partial s}\left( f'_y e^s\sin t \right)$

I'm only going to do one of the above to show you how it's done.

$\displaystyle \frac{\partial }{\partial s}\left( f'_xe^s \cos t \right)=\frac{\partial }{\partial s}\left (f'_{x}\right)e^s\cos t + f'_x\frac{\partial }{\partial s}\left(e^s \cos t \right)=$

$\displaystyle \left \[ f''_{xx} \frac{\partial x}{\partial s} + f''_{xy}\frac{\partial y}{\partial s} \right \]e^s \cos t +f'_x e^s \cos t =$

$\displaystyle f''_{xx}e^{2s}\cos^2 t + f''_{xy}e^{2s}\cos t \sin t + f'_x e^s \cos t$