00:01
In order to find an equation for a tangent line at the point 1 -1, we first have to find the slope for a tangent line.
00:10
And the best way we can do that is through implicit differentiation of our equation x squared plus x y plus y squared equals 3.
00:24
So if we take the derivative, this entire equation, on the left -hand side we'll have the derivative of x squared plus the derivative of x -y plus the derivative of y -square, which will be equal to the derivative of three.
00:49
So starting on the left -hand side, the derivative of x -square is 2x.
00:57
For the derivative of x times y, we will have to use the product rule.
01:02
So we get the derivative of x, which is 1 times y, plus the derivative of y times x.
01:13
So we get x, d y, d x, and then the derivative of y squared, which is 2y, d y, d x.
01:28
And this will be equal to to the derivative of 3, which is 0.
01:33
So now if we subtract 2x and y from both sides of the equation, this leaves us with x times dy -d -x plus 2y times dy -d -x is equal to negative 2x minus y.
01:58
Now we can isolate dydx because if we pull dydx out of this left hand side, we get dydx times x plus 2y, close parenthesis, is equal to negative 2x minus y...