Find the slope of the tangent line to the curve x^2+y^3=5x^2y+5 at the point (1,−1).

Please help!

2 answers

x^2+y^3=5x^2y+5
2x dx+3y^2 dy=10xy dx+5x^2 dy
dx(2x-10xy)=dy(5x^2-3y^2
dy/dx = slope = (5x^2-3y^2)/(2x-10xy) put in 1,-1 and solve
Thank you!! Do you mind checking my other question to see if I did it right?
Similar Questions
  1. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  2. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 3 answers
more similar questions