How do you find #dy/dx# by implicit differentiation given #x^3+y^3=2#?

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Answer 1 · 2017-01-04T14:41:48

I differentiate each term separately, reassemble the equation, and then solve for #dy/dx#

Differentiate the first term:

#(d(x^3))/dx = 3x^2#

Differentiate the second term:

#(d(y^3))/dx = 3y^2dy/dx#

Differentiate the third term:

#(d(2))/dx = 0#

Reassemble the equation:

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

Solve for #dy/dx#:

#3y^2dy/dx = -3x^2#

#dy/dx = -x^2/y^2#