00:02
In this question, we're asked to find the second derivative of y with respect to x at the point one two, given that y is the function of x implicitly defined by this equation that we had in question five.
00:20
Now in that question, we found the derivative of y with respect to x, the first derivative, at the point one two to be negative two over eleven.
00:37
And so we're going to encounter that we're going to need that in this question as well.
00:45
Also in question five, we determined this equation from differentiating once.
00:55
So if we differentiate it again with respect to x, then we can find y double prime.
01:04
Alternatively, we could have differentiated the expression for y prime directly, but then that would also include product, quotient, and chain rules, and also require us to substitute an expression, the expression for y prime, again into the expression, into the right side, which is likely just as much work, if not more, than doing, than differentiating given this equation.
01:39
And so when we do so, we get this for the first term.
01:46
Recall that we need to use chain rule.
01:50
For the second term, we have a triple product rule.
01:55
So that's just, the derivative of this will just be a sum of three terms, where the first term is the derivative, or is just this with this term differentiated, the x...