How do you find the slope of the tangent to the curve #y^3x+y^2x^2=6# at #(2,1)#?

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Answer 1 · 2015-02-20T23:07:43

#3y^2dy/dxx+y^3+2ydy/dxx^2+y^2 2x=0 #

Do some rewriting

#3xy^2dy/dx+2x^2ydy/dx+y^3+2xy^2=0#

Factor and move terms without a #dy/dx # factor to right side

#dy/dx(3xy^2+2x^2y)=-y^3-2xy^2#

now divide both sides by #3xy^2+2x^2y # and factor where you can

#dy/dx=(-y^2(y+2x))/(yx(3y+2x) #

#dy/dx=(-y(y+2x))/(x(3y+3x)) #

Now evaluate at the given point # (2,1)#

#dy/dx=(-1(1+2(2)))/(2(3(1)+2(2)))=(-1(5))/(2(7))=(-5)/14 #