How to find the slope of a tangent line when the X and Ys are multiplied together on the same side?

Question

I have an exam tomorrow and this is one of the problems on the sample questions sheet. I've tried taking d/dx of each variable after putting them on different sides as well as factoring but neither have worked.

The problem is: Find the slope of the tangent to the curve y³x + y²x² = 6 at (2,1) and the answer sheet says that the solution is (-5/14)

If anyone knows how to do this please help!

Thanks so much

Answer 1

"Find the slope of the tangent to the curve y³x + y²x² = 6 at (2,1) and the answer sheet says that the solution is (-5/14)"

Here, you'll want to use a method called implicit differentiation. What happens is that we treat "y" as a constant and by differentiating it with respect to x, we have dy/dx. Our goal is to solve for dy/dx like it is a variable in our equation, and then plug in (2,1) to get the slope of the tangent line:

y³x + y²x² = 6

3y²(dy/dx)x + y³ + 2y(dy/dx)x² + 2xy² = 0 <-- Use Power Rule (works the same w/two variables)

(3xy²+2x²y)(dy/dx) + y³ + 2xy² = 0

y³+2xy² = -(3xy²+2x²y)(dy/dx)

y(y²+2xy) = (-3xy²-2x²y)(dy/dx)

y(y²+2xy) = y(-3xy-2x²)(dy/dx)

(y²+2xy)/(-3xy-2x²) = dy/dx

Now, we plug in (2,1) to get our slope:

dy/dx = [1²+2(2)(1)]/[-3(2)(1)-2(2)²]

dy/dx = (1+4)/[-6-2(4)]

dy/dx = 5/(-6-8)

dy/dx = 5/-14

dy/dx = -5/14

Therefore, the slope of the tangent to the curve y³x + y²x² = 6 at (2,1) is -5/14.

Hope this helped!

Answer 2