How do you determine dy/dx given x^(1/4)+y^(1/4)=2?

1 Answer
Oct 28, 2016

Well, the hardest part is remembering that (dy)/(dy)*(dy)/(dx)=(dy)/(dx).

If you've got this in your head, finding dy/dx should be fairly easy...

To simplify matters though, I will turn 1/4 into the variable n...

This means that:

x^n+y^n=2

Also...

nx^(n-1)+ny^(n-1)*(dy)/(dx)=0

ny^(n-1)*(dy)/(dx)=-nx^(n-1)

(dy)/(dx)=-(nx^(n-1))/(ny^(n-1))

(dy)/(dx)=-(x^(n-1))/(y^(n-1))

(dy)/(dx)=-(x/y)^(n-1)

Now, since n=1/4:

n-1

=1/4-4/4

=-3/4

Which means that:

(dy)/(dx)=-(x/y)^(-3/4)

(dy)/(dx)=-(y/x)^(3/4)

This can be made to look prettier...

(dy)/(dx)=-root(4)((y/x)^3)

And presto!!