# Higher order partial derivatives and the chain rule

Basically, could somebody please explain to me how I find $${\varphi}_u_u$$, I really don't understand how it's come about. Apparantly I need to use the chain rule again and the product rule but I don't understand how to, if somebody could show me explicitly how to use the product rule in this case I'd be very greatful. The bit under the line is the solution but it's not obvious how they get to that. Thanks for any help!

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Anyone?

Here's what I would do:

Take the partial of \phi_{u} with respect to u. Forget about the intermediate step in the solution, I think that required some extra simplification. You care about the end result.

Substitute in for x and y
Pull out of each partial differential any terms that are independent (for example, the partial derivative with respect to u of sin(v)\phi_{x} = sin(v) * partial derivative with respect to u of \phi_{x} )
Then, apply the partial derivatives and substitute back out again for x's and y's. This should get you started, I'm interested to see your work.

arildno
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Here's how you do it very pedantically:
We have:
$$x(u,v)=e^{u}\cos(v),y(u,v)=e^{u}\sin(v)$$
We now define the function $\Phi(u,v)$ as satisfying identically:
$$\Phi(u,v)=\phi(x(u,v),y(u,v))$$

Thus, for example, we get:
$$\Phi_{u}=\phi_{x}x_{u}+\phi_{y}y_{u}$$
And furthermore:
$$Phi_{uu}=\phi_{xx}x_{u}^{2}+\phi_{xy}x_{u}y_{u}+\phi_{yx}x_{u}y_{u}+\phi_{yy}y_{u}^{2}$$
Got that?

Sorry to dig up an old thread, but it turns out I have been set exactly the same question and after spending hours trying to get my head around it, I still can't get the answer they are looking for. I get the same answer as "arildno" which is no the answer we have been asked to show. Can any one help again lol.

got it now :D :D :D
$$\phi_{uu}=(\frac{\partial}{\partial x}(x\phi_{x}+y\phi_{y}))x_{u}+(\frac{\partial}{\partial y}(x\phi_{x}+y\phi_{y}))y_{u}$$

Finally worked out how to use the chain rule correctly. This is the middle step that is missing.