Question

Find dy/dx by implicit differentiation. x^3 + y^3 = 8

Answer 1

First, we derivate both sides of the equation to get the following:

$(x^3+y^3=8)^{\prime}=3x^2+3y^2\cdot y^{\prime}=0$

Notice how the derivative of y^3 has to be written: first you do the derivative using the general polynomial formula and then you write y' as a factor.

Then, we just solve for y' to find dy/dx:

$\begin{gathered} 3x^2+3y^2\cdot y^{\prime}=0 \\ \Rightarrow3y^2\cdot y^{\prime}=-3x^2 \\ \Rightarrow y^{\prime}=\frac{-3x^2}{3y^2}=-\frac{x^2}{y^2} \\ y^{\prime}=-\frac{x^2}{y^2} \end{gathered}$

therefore, the implicit derivative of x^3+y^3=8 is y'=-x^2/y^2