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Assuming that $u = f(x,y)$, $x = e^s\sin(t)$, $y = e^s\sin(t)$

Show that $(\frac{\partial u}{\partial s})^2 \neq \frac{\partial^2 u}{\partial d s^2}$

I know what to do, but I don't know how to do it. The RHS gives me difficulties.

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Were $x$ and $y$ intended to be the same? Is $\sin$ a part of the exponent? How is this discrete? – Karolis Juodelė Nov 14 '12 at 17:01
doesn't discrete math involve proofs? – Arnold Nov 14 '12 at 17:02
All math involves proofs. Discrete is the opposite of continuous. Discrete math involves things like number theory, graphs, etc. – Karolis Juodelė Nov 14 '12 at 17:04
Are you sure is $\frac{\partial^2 u}{\partial ds^s}$ is well written? It doesn't have meaning at all. Also, related: – Pragabhava Nov 14 '12 at 18:04

Since $x=y$, let $f(x,y)=g(x)\\ \frac{\mathrm{d}u}{\mathrm{d}s}=x\frac{\mathrm{d}g}{\mathrm{d}s}\\ \frac{\mathrm{d}^2u}{\mathrm{d}s^2}=\frac{\mathrm{d}g}{\mathrm{d}s}+x\frac{\mathrm{d}^2g}{\mathrm{d}s^2}$

Setting $h=\frac{\mathrm{d}g}{\mathrm{d}s} \ \, \ \frac{\mathrm{d}u}{\mathrm{d}s}=\frac{\mathrm{d}^2u}{\mathrm{d}s^2} \text{iff} \frac{\mathrm{d}h}{\mathrm{d}s}=xh^2-\frac{h}x$. Since a solution must exist, they are unequal only for almost all $u$.

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