INSTRUCTOR: Show that the function f of xy equal to 3x minus 4y squared is differentiable at the point negative 1 comma 2.

We know, by definition, that f of xy, for any function, is equal to f of x0 comma y0 plus fx evaluated at x0 y0 multiplied by, parentheses, x minus x0, close parentheses, plus fy evaluated at x0 y0 multiplied by, open parentheses, y minus y0, close parentheses, plus e of xy, where e of xy is our error polynomial, as it were.

So we know that this is true. What we need to show to prove that this differentiable is that e of xy divided by the square root of-- or x minus negative 1 squared plus y minus 2 or y0 squared. The square root of that big quantity, that ratio goes to 0.

So what we are going to do is find the error polynomial. So we know what f of xy is. And from that, we can find f of negative 1, 2. Evaluating that, we get negative 19.

And we can also find, from this information, fx is equal to 3. And fy is equal to negative 8y. So fx, evaluated at the point negative 1 2 is just 3, f 1, 2, that equals 3. And fy, evaluated at the point negative 1, 2 is equal to negative 16. Those are values we need.

Now, this means that our function f of xy, which is 3x minus 4y squared, is equal to negative 19 plus 3, open parentheses, x plus 1, close parentheses, minus 16, open parentheses, y minus 2, close parentheses, plus e of x comma y.

Solving for e of xy. Rearranging terms, we see that it is equal to negative 4y squared plus 16y minus 16. Having developed this, we want to see-- or show, rather-- that the limit, as xy approaches the point negative 1 comma 2 of e of xy, that would be negative 4y squared plus 16y minus 16 as our numerator, over the denominator, square root x plus y in parentheses squared plus, open parentheses, y minus 2, close parentheses, squared. The square root of all of, that sum, that is equal to 0.

Well, this is going to be equal to the limit as xy approaches the point negative 1 comma 2 of negative 4, open parentheses, y minus 2, close parentheses, squared over the square root of the quantity, parentheses, x plus 1, close parentheses, squared plus, open parentheses, y minus 2, close parentheses, squared.

Which is less than or equal to-- and take some time and convince yourself of this part-- less than or equal to negative 4-- in my numerator-- negative 4, open parentheses, open parentheses again, x plus 1, close parentheses, squared plus, open parentheses, y minus 2, close parentheses, squared, close parentheses, over a denominator of square root of the quantity, open parentheses, x plus 1, close parentheses, squared, plus, open parentheses, y minus 2, close parentheses, squared, the square root of all that. And we can reduce this.

Oh, I left off something important. I said this is less than or equal to that. It's less than or equal to the limit, as xy approaches the point negative 1, 2. That is equal to the limit, as xy approaches the point negative 1 comma 2 of negative 4 square root of the quantity, open parentheses, x plus 1, close parentheses, squared plus, open parentheses, y minus 2, close parentheses, squared. And that limit is equal to 0, which shows that the function f of xy is differentiable at that point negative 1 2.