find the derivative of f(x)=(lnx)^1/x
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calendarEducator since 2012
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f(x)=e^((1/x)ln(ln(x))) since a^b=e^(b ln(a))
the derivative of `ln(ln(x))` is `(1/x)1/(ln(x))`
let `g(x)=(1/x)ln(ln(x))`
`g'(x)=(-1/x^2)ln(ln(x))+(1/x)(1/x)(1/ln(x))`
`g'(x)=-1/x^2 ln(ln(x))+(1/x^2)(1/ln(x))`
`f(x)=e^(g(x))`
`f'(x)=g'(x)e^(g(x))=g'(x)f(x)`
Answer: `f'(x)=(-1/x^2 ln(ln(x))+(1/x^2)(1/ln(x)))ln(x)^(1/x)`
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calendarEducator since 2011
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There are two ways to do this:
First we could write it as
`y = (ln(x))^(1/x)`
`ln(y) = ln(ln(x)^(1/x)) = 1/x ln(ln(x)) = (ln(ln(x)))/x`
Now we can take the derivative of both sides and get
`1/y (dy)/(dx) = (d)/(dx) (ln(ln(x)))/x`
Using the quotient rule we get:
`1/y (dy)/(dx) = ((x(d(ln(ln(x))))/(dx))-(ln(ln(x))))/x^2`
Using the chain rule we can take the derivative of ln(ln(x))
`(d(ln(ln(x))))/(dx) = 1/(ln(x)) (d(ln(x)))/(dx) = 1/(ln(x)) 1/x = 1/(xln(x))`
So we get
`1/y (dy)/(dx) = (x(1/(xln(x)))-ln(ln(x)))/x^2`
`1/y (dy)/(dx) = (1/(ln(x))-ln(ln(x)))/x^2`
Since `y = (ln(ln(x))/x)` we get
` (dy)/(dx) = (1/(ln(x)) - ln(ln(x)))/x^2 * (ln(ln(x)))/x`
`(dy)/(dx) = ((1-ln(x)ln(ln(x)))ln(ln(x)))/(x^3 ln(x))`