How do you find the slope of a tangent line to the graph of the function #x^3 + y^3 – 6xy = 0#, at (4/3, 8/3)?

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Answer 1 · 2018-04-05T03:38:44

#4/5#

#d/dx(x^3+y^3-6xy)=d/dx(0)#

#3x^2+3y^2*(dy)/dx-6(1)y-6x*(dy)/dx=0#

solve for #(dy)/dx#, which gives the tangent slope at point #(x,y)#

#3y^2*(dy)/dx-6x*(dy)/dx=-3x^2+6y#

#(dy)/dx(3y^2-6x)=-3x^2+6y#

#(dy)/dx=(-3x^2+6y)/(3y^2-6x)#

plug in #(4/3,8/3)#:

#(dy)/dx# at #(4/3,8/3) =(-3(4/3)^2+6(8/3))/(3(8/3)^2-6(4/3))#

#(dy)/dx# at #(4/3,8/3) =(-16/3+16)/(64/3-8)#

#(dy)/dx# at #(4/3,8/3) =(32/3)/(40/3)#

#(dy)/dx# at #(4/3,8/3) =4/5#