00:01
Find the derivative of p with respect to x and we're going to use the product rule that state that we need to find the derivative of this first function and we treat this other one which is q x as a constant so this will be two times x minus a and then you know we're going to apply in the chain rule here so we need to find the derivatives of what's inside.
00:30
So it's x minus the derivative of x minus a and that derivative is one.
00:37
And then this doesn't change.
00:39
It's treated as a constant.
00:41
Plus, since we treated qx as a constant, now we're going to treat x minus a squared as a constant.
00:48
So let's write it down x minus a squared and then times the derivative of q of x right and um now we need to evaluate this at x is equal to a okay so we just a substitute with a we see an a so it's going to be a minus a q of x plus a squared times the derivative of q of x this is squared this is a derivative of x this is equals to zero, right? and p of a, we have the equation right here of p of x, but then now we let x equals to a and this will be equals to a minus a squared q of x.
01:41
And this would be equal to zero.
01:44
So this proves that p of a equals to the derivative of p of p of the derivative of p with respect to x, where x is equal to a is equals to 0...