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`...
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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