The complete question is
find the slope of the tangent line and the equation of the tangent line at the given point, [tex]x^2y-5xy^2[/tex][tex]+6=0[/tex] at the point (3,1).
Answer:
Tangent = 1/23 , equation of tangent = 23y - x = 20
Step-by-step explanation:
We are given the equation
[tex]x^2y-5xy^2[/tex][tex]+6=0[/tex]
upon differentiation the given function we get
d ([tex]x^2y-5xy^2[/tex][tex]+6=0[/tex]) /dt
= [tex]2xy +x^2dy/dx - 5y^2 -10xy dy/dx =0[/tex]
upon substituting x=3 ,y=1 we will get
dy/dx = 1/23
now the equation of the tangent will be
y-1 = 1/23 (x-3)
23 (y-1) = x- 3
23y - 23 = x -3
23y - x = 20
