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

##### 2 Answers

#### Explanation:

We will differentiate implicitly. This means, that whenever we differentiate a variable that's not

#d/dxx^3=3x^2" "# but#" "d/dxy^3=3y^2dy/dx# #d/dxx=1" "# but#" "d/dxy=dy/dx# #d/dxxy^2=(d/dxx)y^2+x(d/dxy^2)=y^2+x(2y)dy/dx#

So, for the given example, we see that we will use the power rule on the left-hand side. On the right, don't be misled by the two-thirds power:

#d/dx(x^(2/3)+y^(2/3))=d/dxa^(2/3)#

Power rule:

#2/3x^(-1/3)+2/3y^(-1/3)dy/dx=0#

Multiply both sides by

#x^(-1/3)+y^(-1/3)dy/dx=0#

#1/x^(1/3)+1/y^(1/3)dy/dx=0#

Solving for

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

#dy/dx=-y^(1/3)/x^(1/3)=-(y/x)^(1/3)=-root3(y/x)#

#### Explanation:

Rewrite it