Question

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

Answer 1

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!!