00:02
In this problem, we are asked to find the derivative of the function, y equals x squared minus 4x over x plus 3.
00:13
It's the quotient of two functions, so we are going to use the quotient rule.
00:17
So i'm going to call the function the numerator is u, the function in the denominator is v.
00:24
So the quotient rule tells us that dy over d x is going to be u prime times v minus v prime times u over v square.
00:37
So let's try to find what u prime and v prime are.
00:44
So u prime is going to be d u over d x.
00:50
So that's going to be d over d x of x squared minus 4x.
00:59
Now derivative of x square by using power rule is going to be 2x, derivative of 4x.
01:05
We pull out the constant 4 and then derivative of x is just 4.
01:10
So this derivative is 2x minus 4.
01:15
Similarly, v prime is dv over dx.
01:20
So that's d over d x of x plus 3.
01:26
And we know that derivative of a sum is the sum of derivatives...