00:01
So for this problem, we want to find d .ydx at the point to 1.
00:05
And so we can go ahead and start by justing the derivative of our function with respect to x.
00:11
So we know that we're going to need to use the quotient rule here.
00:16
So the question rule tells us that we're going to multiply our denominator, which is x minus y, by the derivative d, d, d, x, plus y, minus the numerator x plus y, times the derivative d d d x of our denominator x minus y over our denominator squared so over x minus y squared and this is going to be equal to the derivative of a constant three which is just zero um so we can go ahead and start taking the derivative here um so our first term is going to be x minus y times d d d d x plus is going to be 1 plus the derivative of y with respect to x is just d y d x minus x plus y again times d d x or d d d x of x minus y d x minus y d x um is going to be one minus d y d x um and we can actually just go ahead and multiply both sides of our equation by x minus y squared um so essentially that just gets rid of it because it's going to be multiplied by 0 um and from here we're going to want to distribute out each of our terms.
01:21
So we get x plus x, d, y, d, x, minus y, d, x, for our first set of parentheses.
01:34
And then we get minus x minus x d y d y d x plus y d y d x, that is equal to zero.
01:46
And we're going to go ahead and write all of our d y d x terms on the opposite side of our equation...