How to solve this?

Question

Find the equations of tangents for the implicit function x³y+cos(x²y)=x² at the points on the curve where y=0.

Answer 1

Let's first calculate all the points on the curve where y = 0.

Substitute 0 for y:

cos(0) = x²

1 = x²

Therefore there are just two points:

(-1, 0), (1, 0)

To find the slope of the tangents, we need dy/dx.

Differentiate both sides of the given equation with respect to x.

2x²y + x³dy/dx - sin(x²y)(2xy + x²dy/dx) = 2x

Substitute y=0 into the above result

x³dy/dx = 2x

dy/dx = 2/x²

Thus at both points (-1, 0), (1, 0)

dy/dx = 2

So the two tangents are

y = 2x + 2

y = 2x - 2