How do you find #dy/dx# given #x=3y^(1/3)+2y#?

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Answer 1 · 2017-01-24T12:11:53

#dy/dx=y^(2/3)/(1+2y^(2/3))#

We are differentiating with respect to #x#. This means that #d/dxx=1#, but #d/dxy=dy/dx#.

This will be recurrent throughout this problem: differentiating anything with #y#, since it's a function that's not #x#, will take the chain rule.

Thus, the derivative with respect to #x# of #y^2# is not just #2y#, but #2ydy/dx#.

Now proceeding my differentiating:

#d/dxx=3d/dxy^(1/3)+2d/dxy#

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

Factoring:

#dy/dx(y^(-2/3)+2)=1#

#dy/dx=1/(y^(-2/3)+2)#

#dy/dx=y^(2/3)/(1+2y^(2/3))#