If u=f(r), where r^2=x^2+y^2, then left(frac{ | 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 $\fr

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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, $u_{xx} = \frac{\partial}{\partial x} \left( f^{\prime}(r) \frac{x}{r} \right) = f^{\prime \prime}(r) \frac{x^2}{r^2} + f^{\prime}(r) \left( \frac{r - x(x/r)}{r^2} \right) = f^{\prime \prime}(r) \frac{x^2}{r^2} + f^{\prime}(r) \frac{r^2 - x^2}{r^3}$. Similarly, $u_{yy} = f^{\prime \prime}(r) \frac{y^2}{r^2} + f^{\prime}(r) \frac{r^2 - y^2}{r^3}$. Adding these: $u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{x^2+y^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - (x^2+y^2)}{r^3} \right)$. Since $x^2+y^2 = r^2$: $u_{xx} + u_{yy} = f^{\prime \prime}(r) \left( \frac{r^2}{r^2} \right) + f^{\prime}(r) \left( \frac{2r^2 - r^2}{r^3} \right) = f^{\prime \prime}(r) + f^{\prime}(r) \frac{r^2}{r^3} = f^{\prime \prime}(r) + \frac{1}{r} f^{\prime}(r)$.

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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