How do you differentiate y=x^(2/x)?

2 Answers
Nov 17, 2016

-4/x^3x^((2-x)/x)

Explanation:

Apply Chain rule of differentiation, with u=2/x:
y'=(x^u)'\times(2/x)'
set differentiation \rArr(ux^(u-1))\times-2(x)^(-2)
apply u=(2/x): \rArr(2/x)x^(2/x-1)\times-2x^(-2)
fraction form of -2x^-2: \rArr-(2/x)x^((2-x)/x)2/x^2

  • the two fractions can multiply by each other.
  • so your answer is...

-4/x^3x^((2-x)/x)

Nov 17, 2016

(dy)/(dx)=2(1-lnx)x^(2/x-2)

Explanation:

As y=x^(2/x)

we have lny=2/xlnx

or 2lnx=xlny

Now deriving both sides w.r.t. x

2/x=lny+x/y*(dy)/(dx)

or (dy)/(dx)=(2/x-lny)*y/x

= (2/x-2/xlnx)*y/x

= 2/x(1-lnx)x^(2/x)/x

= 2(1-lnx)x^(2/x-2)