00:02
In this problem, we want to find all points on the given curve where the slope of a tangent line is equal to minus 1.
00:15
So essentially, what they're asking us is to find coordinates xy such that d y over the x at these coordinates is equal to minus 1.
00:37
This is what it means for the slope of a tangent line to be equal to minus 1, because our derivative essentially will give us a slope over the tangent line of our curve.
00:50
But how do we find the y over the x when we are given an implicit function? an implicit function means that we can't write y as a function of x.
01:02
We can't solve for y, but we still want to find the y over d x, and how we're going to do that is we're going to differentiate implicitly.
01:11
Meaning that we're going to apply the derivative with respect to x on both the left -hand side of our equation and the right -hand side of our equation.
01:28
Developing our left -hand side of the equation, we'll find that this would give us two smaller derivatives to evaluate.
01:35
D over the x of x squared times y -square, plus d over the x of x times y.
01:47
And this will be equal to the derivative of 2.
01:51
But since 2 is a constant, its derivative is equal to 0.
01:57
So let's develop our left -hand side by evaluating our derivatives.
02:03
Here, the derivative x squared times y square will require use of product rule.
02:09
So this will give us 2 times x times y square plus 2 times x squared plus 2 times x squared times y dy over d x this d y over d x comes about due to chain rule now let's do the same for the second derivative which we will also differentiate via product rule this would give us y plus x times d y over d x so now let's solve for d y over d x and factor it out so the y over d x times 2x squared times y x plus x would be equal to minus 2x y square minus y so now we can write that the y over the x is equal to minus two times x times y over two times x square y plus x oh excuse me this is a plus here because i fact about the minus sign so now in our numerator, let's factor out a factor of y, and in our denominator, we'll factor out 1x.
03:56
This will give us, in brackets, the factor 2xy plus 1 over 2xy plus 1.
04:15
So here we have that a term in bracket actually cancels out and is equal to 1.
04:21
Implying that the y over d x is actually equal to minus x over y...