00:01
Okay, so to solve this, we want to show that if the derivative f prime is equal to a constant k times f of x, then the function itself is a constant c times e to the kx.
00:13
Now, let's first recall what we've already shown previously in this section is that if the derivative is equal to itself, f prime is equal to f, then the function is equal to c times e to x.
00:26
So in other words, we've really shown this is true for k equals 1, but we want to show it for general k.
00:34
And to do that, we use a similar technique that is done previously in this section, which is to take the ratio of the two things, f of x and e to the kx, and show that this derivative of this quotient is zero.
00:53
So then we'll see how this gets us.
00:56
To do this, we use the quotient rule, it's a quotient of two functions.
01:00
So it's going to be bottom function times the derivative of the top minus the top function times the derivative of the bottom.
01:09
The derivative of the bottom, again, is this by the derivative of exponentials, it's divided by the bottom function squared...