00:01
Define what the derivative of g equals when we set it equal to zero.
00:05
We first need to find what g of x equals when we set it to zero.
00:10
Let's start by setting this equation right here in for zero.
00:16
If we plug in a zero into g, we don't know what it is because it's the function.
00:22
We don't know what it is.
00:23
Plus, plug in zero for x, it's going to be times zero.
00:29
Sign of g to the zero is equal to zero.
00:37
Squared.
00:38
Okay, now we can solve this out to find what g of zero equals.
00:46
This is going to cancel out because it's multiplied by zero and this is also just equal zero because zero squared is zero.
00:52
So this means g of zero is also equal to zero.
01:01
Now to find the derivative of g for zero, we can rewrite our equation in terms of y and x.
01:11
G of x will equal y and the derivative of g of x will equal y prime or our derivative of y.
01:21
I'll rewrite it right here.
01:37
Okay, now we can solve our new equation using implicit differentiation.
01:40
We can start by differentiating both sides as the equation in respect to x.
01:55
Okay, i'll sort of the left side.
01:57
The derivative of y is just going to be one, but since we're differentiating in respect to x, it's going to be y prime.
02:05
And then we have a product rule.
02:07
We'll start with a derivative of x, which is just 1.
02:09
So it's going to be just sine y, plus our derivative of sine of y, which is cosine of y...