How do you determine dy/dx given #x^(1/3)+y^(1/3)=7#?

1 Answer
May 13, 2017

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

Explanation:

Note that #d/dxx^(1/3)=1/3x^(-2/3)# through the power rule.

The derivative with respect to #x# of #y^(1/3)# is found identically, except for that since #y# is its own function, the chain rule will come into effect, and we will have to multiply the derivative found through the product rule by the derivative of #y#, which is #dy/dx#.

That is #d/dxy^(1/3)=1/3y^(-2/3)*dy/dx#.

So we have:

#x^(1/3)+y^(1/3)=7#

And taking the derivative of both sides:

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

Multiplying both sides by #3# and rewriting:

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

Solving for #dy/dx#:

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

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