Question

prove that del^2f(r) = f"(r) + (2/r) f'(r) where r = sqrt(x^2+y^2+z^2)

Answer 1

Answer:

Proved

Step-by-step explanation:

Step 1: For calculating the derivative of the composition of two or more functions, use the Chain Rule formula. For composite functions, the differentiation chain rule is defined. The chain rule, for instance, describes the derivative of the composite of two functions, f and g, if they are functions.

Step 2: By using the chain rule, we have

$\frac{\partial f}{\partial x}=\frac{d f}{d r} \cdot \frac{\partial r}{\partial x}$

and

$\begin{aligned}\frac{\partial^2 f}{\partial x^2} & =\frac{\partial}{\partial x}\left(\frac{d f}{d r}\right) \cdot \frac{\partial r}{\partial x}+\frac{d f}{d r} \cdot \frac{\partial^2 r}{\partial x^2} \\& =\frac{\partial^2 f}{\partial r^2} \cdot\left(\frac{\partial r}{\partial x}\right)^2+\frac{d f}{d r} \cdot \frac{\partial^2 r}{\partial x^2} \\& =\frac{\partial^2 f}{\partial r^2} \cdot\left(\frac{x}{r}\right)^2+\frac{d f}{d r} \cdot\left(\frac{1}{r}-\frac{x^2}{r^3}\right)\end{aligned}$

Thus,

$\begin{aligned}\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2} & =\frac{\partial^2 f}{\partial r^2} \cdot\left[\left(\frac{x}{r}\right)^2+\left(\frac{y}{r}\right)^2+\left(\frac{z}{r}\right)^2\right] \\& +\frac{d f}{d r} \cdot\left(\frac{3}{r}-\frac{x^2+y^2+z^2}{r^3}\right) \\& =\frac{\partial^2 f}{\partial r^2}+\frac{2}{r} \frac{d f}{d r}\end{aligned}$

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