How do you use implicit differentiation to find the slope of the curve given #1/x^3+1/y^3=2# at (2,2)?

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Answer 1 · 2016-11-15T02:36:44

This is either a 'trick' question or there is an error. The point #(2,2)# is not on the graph of #1/x^3+1/y^3=2#

We can find #dy/dx# using implicit differentiation.

#x^-3+y^-3 = 2# (or some other constant. If we use #1/4# then #(2,2)# is on the graph.)

Differentiate both sides with respect to #x#:

#-3x^-4-3y^-4 dy/dx = 0#

Solve for #dy/dx#

#-3/x^4-3/y^4 dy/dx = 0#

#dy/dx = -y^4/x^4#

At any point #(a,a)# that is on the graph, the slope of the tangent line is #-1#.