# I'm having major difficulties with partial differentiation using the chain rule

1. Feb 1, 2012

### s3a

This is another problem than I've been stuck on for a long time and I tried reading and watching videos but I only find first order partial differentiation with more than two variables or higher order partial differentiation with only two variables. (I'm not calling f a variable but I am calling x and y variables even though they are functions of other variables.)

I am stuck very close to the beginning and am attaching my work as stuck.jpg and have the solution to the problem (which is uploaded as solution.jpg) but I do not understand how to get to u_(ss). Could someone show me how to apply the tree diagram to that?

Any input would be GREATLY appreciated!

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2. Feb 1, 2012

### susskind_leon

Do you understand how to obtain u_s? I think that's the more crucial part. Once you know how to get to u_s you should be able to get to u_ss. Just keep in mind that f=f(x,y) as well as f_x=f_x(x,y) and f_y=f_y(x,y). And keep in mind the product rule as well.

3. Feb 1, 2012

### s3a

I do get how to get u_s and I know I should be able to know how to get u_(ss) from that but for some reason I don't :(.

I also know that f = f(x,y) as well as f_x = f_x(x,y) and f_y = f_y(x,y) and I vaguely suspected that the product rule was being applied for the u_(ss) step but can you outline the steps from u_s to u_(ss) please?

4. Feb 1, 2012

### SammyS

Staff Emeritus

You have
$\displaystyle \frac{\partial u}{\partial s}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$

$\displaystyle u_s= f_x\,\frac{\partial x}{\partial s}+f_y\,\frac{\partial y}{\partial s}= f_x\,e^s\cos(t)+f_y\,e^s\sin(t)$​
So, for uss take the partial of us w.r.t. s, especially observing the chain rule & product rule.

First the product rule:
$\displaystyle \frac{\partial u_s}{\partial s} = \frac{\partial f_x}{\partial s} \frac{\partial x}{\partial s}+f_x\,\frac{\partial^2 x}{\partial s^2}+ \frac{\partial f_y}{\partial s} \frac{\partial y}{\partial s}+f_y\,\frac{\partial^2 y}{\partial s^2}$​

Now plug-in the chain rule for $\displaystyle \frac{\partial f_x}{\partial s}\text{ and }\frac{\partial f_y}{\partial s}\,.$

$\displaystyle \frac{\partial u_s}{\partial s} =\left(\frac{\partial f_x}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f_x}{\partial y}\frac{\partial y}{\partial s}\right) \frac{\partial x}{\partial s}+f_x\,\frac{\partial^2 x}{\partial s^2}+ \left(\frac{\partial f_y}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f_y}{\partial y}\frac{\partial y}{\partial s}\right) \frac{\partial y}{\partial s}+f_y\,\frac{\partial^2 y}{\partial s^2}$
The rest is algebra.