`y = x^(2/x)` Use logarithmic differentiation to find dy/dx

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marizi eNotes educator| Certified Educator

 For the given problem:` y = x^(2/x)` , we apply the natural logarithm on both sides:

`ln(y) =ln(x^(2/x))`

Apply the natural logarithm property: `ln(x^n) = n*ln(x)` .

`ln(y) = (2/x) *ln(x)`

Apply chain rule  on the left side since y is is function of x.

`d/dx(ln(y))= 1/y *y'`


Apply product rule:` d/(dx) (u*v) = u'*v + v' *u` on the right side:

Let `u=2/x` then `u' = -2/x^2`

    ` v =ln(x)` then` v' = 1/x`

`d/(dx) ((2/x) *ln(x)) =d/(dx) ((2/x)) *ln(x) +(2/x) *d/(dx) (ln(x))`

                                `= (-2/x^2)*ln(x) + (2/x)(1/x)`

                              ` =(-2)/(x^2ln(x))+ 2/x^2`

                          ` = (-2ln(x)+2)/x^2`


The derivative of `ln(y) = (2/x) *ln(x) ` becomes :


 Isolate y' by multiplying both sides by (y):

`y* (1/y*y')= ((-2ln(x)+2)/x^2)*y`

`y' =((-2ln(x)+2)*y)/x^2`

Plug-in `y = x^(2/x) `  on the right side:

`y' =((-2ln(x)+2)*x^(2/x))/x^2`


Or `y' =((-2ln(x)+2)*x^(2/x))*x^(-2)`

   `y' =(-2ln(x)+2)*x^(2/x-2)`

   `y' =(-2ln(x)+2)*x^((2-2x)/x)`

   ` y' =-2x^((2-2x)/x)ln(x)+2x^((2-2x)/x)`

`    y = -2x^((2-2x)/x) (lnx-1)