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

##### 1 Answer

Jan 24, 2017

#### Explanation:

We are differentiating *with respect to*

This will be recurrent throughout this problem: differentiating anything with

Thus, the derivative with respect to

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))#