I have a question about partial differential equations

Question

How to find $\frac{\partial^2 e^u}{\partial y^2} =$? When $u = u(x,y)$

Answer 1

$\frac{\partial e^u}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial e^u}{\partial u}\frac{du}{dy}\right) = e^u \left[\frac{\partial^2 u}{\partial y^2} + \left(\frac{\partial u}{\partial y}\right)^2\right]$

Answer 2

Here $~u = u(x,y)$ and let $~f(u)=e^u$.

So by chain rule, $$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial y} $$and

$$\frac{\partial^2 f}{\partial y^2} =\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y} \right) =\frac{\partial }{\partial y} \left(\frac{\partial f}{\partial u}\cdot\frac{\partial u}{\partial y} \right)=\frac{\partial f}{\partial u}\cdot\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 f}{\partial u^2}\cdot \left(\frac{\partial u}{\partial y} \right)^2$$ So $$\frac{\partial^2 e^u}{\partial y^2}= \frac{\partial e^u}{\partial u}\cdot\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 e^u}{\partial u^2}\cdot \left(\frac{\partial u}{\partial y} \right)^2$$ $$\implies \frac{\partial^2 e^u}{\partial y^2}=e^u\left\{\frac{\partial^2 u}{\partial y^2}+\left(\frac{\partial u}{\partial y}\right)^2\right\}$$

Answer 3

$$\frac{\partial}{\partial y}e^u=e^u\frac{\partial u}{\partial y},$$

$$\frac{\partial^2}{\partial y^2}e^u =\frac{\partial}{\partial y}\left(e^u\frac{\partial u}{\partial y}\right) =\left(\frac{\partial}{\partial y}e^u\right)\frac{\partial u}{\partial y}+e^u\frac{\partial}{\partial y}\frac{\partial u}{\partial y} =e^u\left(\frac{\partial u}{\partial y}\right)^2+e^u\frac{\partial^2u}{\partial y^2}.$$

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Or more concisely,

$$\frac{\partial}{\partial y}e^u=e^uu_y,$$

$$\frac{\partial^2}{\partial y^2}e^u=\frac{\partial}{\partial y}\left(e^uu_y\right)=\left(\frac{\partial}{\partial y}e^u\right)u_y+e^u\frac{\partial}{\partial y}u_y=e^uu_y^2+e^uu_{yy}.$$