Calculus homework help with tangent line slope?

Question

the point (1,-1) lies on the curve x^3y^3+x+y^2=1 
. 
Find the slope of the tangent line to the curve at this point

Answer 1

The curve is  x³y³ + x + y² = 1 and the point is (1, - 1).

x³y³ + x + y² = 1.

Differentiate the curve with respect to x.

Use product rule differentiation formula : (uv) ' = uv ' + vu '.

Use power rule of differentiation formula : (xⁿ) ' = nx^{n - 1}.

x³(3y²)y ' + y³(3x²) + 1 + 2yy ' = 0

3x³y²y ' + 3x²y³ + 1 + 2yy ' = 0

y '(3x³y² + 2y) + 3x²y³ + 1 = 0

y ' = - [3x²y³ + 1] /[3x³y² + 2y]

At the point (1, - 1), y ' = - [3(1)²(- 1)³ + 1] / [3(1)³(- 1)² + 2(- 1)]

= - [- 3 + 1] / [3 - 2]

= 2 / 1

= 2.

⇒ y ' = 2.

This is the slope (m ) of the tangent line to the curve at (1, - 1).