Let u=f(r) and r=|x|=(x1^2 +···|xn^2)^1/2.
Show that ∑ 1 to n (∂u/∂xj)^2 = (f′(r))^2
To show this, we will first calculate the partial derivative of u with respect to xj. Then we will square the obtained derivative and sum it over j from 1 to n.
1. Calculate the partial derivative of u with respect to xj:
Since u = f(r), we can calculate the derivative using the chain rule. The chain rule states that if u = f(g(x)), then the derivative du/dx is given by du/dx = f′(g(x)) * g′(x).
In our case, u = f(r) and r = √(x1^2 + x2^2 + ... + xn^2). Applying the chain rule, we have:
∂u/∂xj = f′(r) * ∂r/∂xj
Now, let's calculate ∂r/∂xj:
We can express r as r = √(x1^2 + x2^2 + ... + xj^2 + ... + xn^2). Deriving this with respect to xj, we get:
∂r/∂xj = (1/2) * (x1^2 + x2^2 + ... + xj^2 + ... + xn^2)^(-1/2) * 2xj = xj / r
Substituting ∂r/∂xj = xj / r into the partial derivative of u, we have:
∂u/∂xj = f′(r) * (xj / r)
2. Square the obtained derivative and sum it over j from 1 to n:
(∂u/∂xj)^2 = [f′(r) * (xj / r)]^2 = (f′(r))^2 * (xj / r)^2 = (f′(r))^2 * xj^2 / r^2
Now we sum (∂u/∂xj)^2 over j from 1 to n:
∑[(∂u/∂xj)^2] = ∑[(f′(r))^2 * xj^2 / r^2]
Since xj^2 are constant with respect to the sum, we can take them out:
∑[(∂u/∂xj)^2] = (f′(r))^2 * (1/r^2) * ∑[xj^2]
The sum ∑[xj^2] is simply the sum of squared components of a vector, which equals r^2:
∑[(∂u/∂xj)^2] = (f′(r))^2 * (1/r^2) * r^2 = (f′(r))^2
Therefore, we have proven that ∑[∂u/∂xj)^2 = (f′(r))^2.