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

 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 :
1/y*y'=(-2ln(x)+2)/x^2
 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)

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