00:01
For this problem, we'll have to calculate the second derivative of t over t plus 1, and then evaluate a t as equal 1.
00:09
So the main thing here is we can't break apart the term, and so we're either going to have to do quotient rule or product rule twice.
00:18
And we can do product rule by moving the t plus 1 to the top, making it a negative 1.
00:26
Considering that this is a prime quotient rule example, i will be using quotient rule for this.
00:33
So the quotient rule by definition is the low function, so that is the function on bottom, times the derivative of the high function.
00:44
So i remember this as low, d high, minus high, d low, over low square.
00:54
So we're going to need our low.
00:58
Well, our low is t plus one, and then our high is t.
01:04
Our derivative of our low is one, and our derivative of our high, in this case, is one.
01:12
So putting all of this into our equation, we're going to have t plus one times one minus t times one over t plus one squared.
01:26
And the t's on top will cancel.
01:30
We'll be off with just one over t plus one squared.
01:34
And so that's our first derivative.
01:36
But we need our second derivative, so we're going to do the same thing.
01:39
So in this case, the top is one.
01:41
So our high is going to be one.
01:45
The derivative of our high would be zero...