00:01
Let's talk about problem 8.
00:03
This one says that x cubed is equal to 2x minus y over 1 minus x square.
00:15
Let's differentiate both sides.
00:17
So we have d over d x of x cube and this will be equal to d over d x of 2x minus y over 1 minus x square.
00:28
Over to the left we'll get 3x square and over to the right we have to use the quotient rule.
00:34
So in quotient should we have denominator square, then we have denominator over here, then we'll differentiate the numerator.
00:46
So this is 2x minus y, minus numerator, which is 2x minus y, and then we differentiate the denominator, which is 1 minus x square.
00:59
So this further simplifies to 3x square will be the denominator remains as it is.
01:06
So this is 1 minus x square whole square.
01:10
We have 1 minus x square over here.
01:12
We take this differentiation inside...