Let u=f(r) and r=|x|=((x1)^2 +···|(xn)^2)^1/2.
Show that ∑ (∂u/∂xj)^2 = (f′(r))^2
Please help! it's an assignment due tomorrow
Think about this:
u^2 = r^2
2u u' = 2r r'
u^2 = ∑ xj^2
2u du = 2∑ xj
To prove the given equation, we need to start by finding the partial derivative of u with respect to xj (denoted as ∂u/∂xj).
First, let's simplify the expression for r:
r = |x| = (x1^2 + x2^2 + ... + xn^2)^1/2
Next, applying the chain rule, we find that:
∂u/∂xj = ∂f/∂r * ∂r/∂xj
Using the chain rule, the derivative of f(r) with respect to r, denoted as f'(r), is:
f'(r) = ∂f/∂r
Now, let's find the derivative of r with respect to xj:
∂r/∂xj = 1/2 * (x1^2 + x2^2 + ... + xn^2)^(-1/2) * 2xj
= xj / r
Substituting the derivatives back into our expression for ∂u/∂xj:
∂u/∂xj = ∂f/∂r * ∂r/∂xj
= f'(r) * (xj / r)
Taking the square of this expression, we have:
(∂u/∂xj)^2 = (f'(r))^2 * (xj / r)^2
= (f'(r))^2 * (xj^2 / r^2)
Now, we need to calculate the sum of these squared partial derivatives (∑ (∂u/∂xj)^2). Let's do that:
∑ (∂u/∂xj)^2 = ∑ [(f'(r))^2 * (xj^2 / r^2)]
Since the summation symbol ∑ indicates that we're summing over all j, we can take (f'(r))^2 and (xj^2 / r^2) outside the summation:
∑ (∂u/∂xj)^2 = (f'(r))^2 * ∑ (xj^2 / r^2)
The summation ∑ (xj^2 / r^2) is equivalent to 1, since we are summing over all j and both xj^2 and r^2 will be included. Therefore, we can simplify the equation further:
∑ (∂u/∂xj)^2 = (f'(r))^2 * 1
Finally, we have:
∑ (∂u/∂xj)^2 = (f'(r))^2
Hence, we have proved that ∑ (∂u/∂xj)^2 = (f'(r))^2.