00:01
So for this question, we're given some values of u and u prime at the location zero, and we're asked to find the derivative of these functions at the location zero.
00:12
So basically this is just testing our various derivation rules.
00:17
So for example, for uv, we know that the derivative is going to be equal to u prime v plus b prime u and so in that case that would be u prime which is minus three times minus one or three plus v prime which is two times five is ten or thirteen and then here we have a quotient rule, so we know that we're going to be looking for v -u -prime minus u -b -prime all over v squared.
01:08
And so that will give us a, so v is minus 1 times u -prime is minus 3.
01:16
So that's a 3 minus 10 all over b squared, which will just be 1.
01:24
And so that will give us a negative 7.
01:27
Then this time we're doing the same thing but sort of reverse here.
01:31
So this will be a u squared.
01:33
And this time we'll have u b prime minus b u prime...