What is the slope of the tangent line of #(x-y)^3-y^2-yx= C #, where C is an arbitrary constant, at #(1,-1)#?

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Answer 1 · 2016-11-04T17:23:43

slope = #dy/dx=13/11#

To find the slope of the tangent line we have to find the derivative of the function first.

#(x-y)^3-y^2-yx=C#

#3(x-y)^2-3(x-y)^2 dy/dx-2y dy/dx-y-xdy/dx=0#

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

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

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

Now plug in 1 for x and -1 for y

#dy/dx=(3(1--1)^2--1)/(3(1--1)^2+2(-1)+1#

#dy/dx=13/11#