What is the slope of the curve #6x^3 + 4y^3 = 6# at the point #(3, -3)#?

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Answer 1 · 2015-11-19T18:16:21

The point #(3,-3)# is not on that curve, but it is on #6x^3+4y^3=54#.
The slope at #(3,-3)# is #-3/2#

#6x^3+4y^3=54#

#d/dx(6x^3+4y^3) = d/dx(54)#

#18x^2+12y^2 dy/dx = 0#

At #(3,-3)#, we get:

#18(9)+12(9)dy/dx = 0#

So,

#dy/dx = (-18(9))/(12(9)) = =18/12 = -3/2#