00:01
In this question, we're asked to find two functions, f and g, functions of x, so that the derivative of f over g is simply f prime over g prime.
00:09
In general, we know this isn't true.
00:11
We have to use a quotient rule.
00:13
We can actually find certain functions so that it is true.
00:18
Let's use the functions f is equal to x over 1 minus x, and g is equal to x.
00:29
So let's prove that once you plug these in to the top equation, that it's true.
00:38
Let's look at the left side first.
00:43
Let's just first calculate what f over g is before we take the derivative.
00:48
So x over 1 minus x divided by x.
00:55
We're going to rewrite this so we actually multiply by the reciprocal instead of dividing.
01:04
Once the x is canceled, we get nicely 1 over 1 minus x.
01:09
Now we can take the derivative of 1 minus 1 minus x.
01:22
We'll use the quotient rule, so low, lower function, times the derivative of the higher function, which is 0, minus the higher function, times the derivative of the lower, which is negative 1.
01:36
Then we just multiply by the low function squared, 1 minus x squared...