If u=f(r),where r^2=x^2+y^2,then left(frac{pa | vedclass

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(C) Given $u=f(r)$ and $r^2=x^2+y^2$. First,find the partial derivatives with respect to $x$: $\frac{\partial r}{\partial x} = \frac{x}{r}$ and $\frac

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$f^{\prime \prime}(r)+\frac{1}{r} f^{\prime}(r)$

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(C) Given $u=f(r)$ and $r^2=x^2+y^2$. First,find the partial derivatives with respect to $x$: $\frac{\partial r}{\partial x} = \frac{x}{r}$ and $\frac{\partial r}{\partial y} = \frac{y}{r}$. Using the chain rule,$u_x = f^{\prime}(r) \frac{\partial r}{\partial x} = f^{\prime}(r) \frac{x}{r}$. Then,$u_{xx} = \frac{\partial}{\partial x} \left( f^{\prime}(r) \frac{x}{r} \right) = f^{\prime \prime}(r) \left( \frac{x}{r} \right)^2 + f^{\prime}(r) \left( \frac{r - x \frac{\partial r}{\partial x}}{r^2} \right) = f^{\prime \prime}(r) \frac{x^2}{r^2} + f^{\prime}(r) \frac{r^2 - x^2}{r^3}$. Similarly,$u_{yy} = f^{\prime \prime}(r) \frac{y^2}{r^2} + f^{\prime}(r) \frac{r^2 - y^2}{r^3}$. Adding these two: $u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{x^2+y^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - (x^2+y^2)}{r^3} \right)$. Since $x^2+y^2 = r^2$,we get: $u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{r^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - r^2}{r^3} \right) = f^{\prime \prime}(r) + f^{\prime}(r) \frac{r^2}{r^3} = f^{\prime \prime}(r) + \frac{1}{r} f^{\prime}(r)$.

If $u=f(r)$,where $r^2=x^2+y^2$,then $\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)$ is equal to

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