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!!