00:01
The question gives us the values of f at 5, f prime at 5, g at 5, and g prime at 5, and asks us to solve for the derivative of f times g at 5, the derivative of f over g at 5, and the derivative of g over f at 5.
00:15
So to start at a, we can see that the derivative of f times g is just our product rule, and we know that equals f prime g plus g prime f.
00:29
And now if we want the value at 5, we need f prime at 5 times g at 5 plus g prime at 5 plus times f at 5.
00:54
And now again, the problem gives us the values of f prime at 5, which is 6.
01:02
We multiply that by g at 5 is negative 3, add that to g prime.
01:08
At 5 is 2 and multiply it by f at 5 is 1 and we can simplify we get negative 18 plus 2 and this is negative 16 which is our answer for part a and now for part e b we can see that the derivative of f over g takes the form of the quotient rule which we know is f prime g minus g prime f over g squared.
01:47
And again, if we want the value at 5, we need f prime at 5 times g at 5 minus g prime at 5 times f at 5 over g at 5 squared.
02:12
And so we can plug in the values that the problem gives us.
02:16
F prime at 5 is 6, times g at 5 is negative 3 minus g prime at 5 is 2 times f at 5 is 1 over g at 5 squared, which is negative 3 squared.
02:34
And if we simplify, we get negative 18 minus 2 over 9.
02:41
And then we can come to our answer, negative 20 over 9...