00:01
So here we're asked to apply derivative rules.
00:04
The difference rule will help us in the first, which is very much like the sum rule.
00:11
If you have f minus g and when it's derivative, you can simply find their separate derivatives and subtract.
00:17
So if we want this evaluated at 3, we'll plug in 3 to f prime, plug in 3 to g prime, and subtract those values.
00:26
Well, we're given the f prime of 3 is 2, g prime of 3 is negative 10, upon subtraction, we get 12.
00:36
For b, we need to use the product rule.
00:39
Let's recall what that is.
00:41
So f times g prime, we'll take the derivative of the first, but leave the second unchanged, and then just switch those roles of what's differentiated and what's left alone.
00:52
So we have f times g prime at three in particular, would be just each function evaluated at three.
01:02
And we'll plug in what we know.
01:04
F prime of three is two, g of three is six, 6, g prime of 3 is negative 10, and f of 3 is 7...