Question

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

Answer 1

`-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)`

Answer 2

`(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)`