Asked by sal

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

Please help!
Answered by bobpursley
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
Answered by sal
Thank you!! Do you mind checking my other question to see if I did it right?
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