What is the derivative of (ln(x)^{1/x}) ?

The derivative of (ln(x)^{1/x}) is ln^{1/x}(x)(-(ln(ln(x)))/(x^2)+1/(x^2ln(x)))

What is the first derivative of (ln(x)^{1/x}) ?

The first derivative of (ln(x)^{1/x}) is ln^{1/x}(x)(-(ln(ln(x)))/(x^2)+1/(x^2ln(x)))

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\frac{d}{d x} ((ln (x))^{\frac{1}{x}})

ln^{\frac{1}{x}} (x) (- \frac{ln ( ln ( x ) )}{x ^{2}} + \frac{1}{x ^{2} ln ( x )})

Solution steps

\frac{d}{d x} ((ln (x))^{\frac{1}{x}})

Apply exponent rule:

a^{b} = e^{b l n (a)}

(ln (x))^{\frac{1}{x}} = e^{\frac{1}{x} l n (l n (x))}

= \frac{d}{d x} (e^{\frac{1}{x} l n (l n (x))})

Apply the chain rule:

e^{\frac{1}{x} l n (l n (x))} \frac{d}{d x} (\frac{1}{x} ln (ln (x)))

= e^{\frac{1}{x} l n (l n (x))} \frac{d}{d x} (\frac{1}{x} ln (ln (x)))

\frac{d}{d x} (\frac{1}{x} ln (ln (x))) = - \frac{ln ( ln ( x ) )}{x ^{2}} + \frac{1}{x ^{2} ln ( x )}

= e^{\frac{1}{x} l n (l n (x))} (- \frac{ln ( ln ( x ) )}{x ^{2}} + \frac{1}{x ^{2} ln ( x )})

e^{\frac{1}{x} l n (l n (x))} = ln^{\frac{1}{x}} (x)

= ln^{\frac{1}{x}} (x) (- \frac{ln ( ln ( x ) )}{x ^{2}} + \frac{1}{x ^{2} ln ( x )})

\int_{0}^{1} (x - \frac{1}{2})^{2} d x

\frac{d y}{d t} - 2 y = 3 e^{- 2 t}

\frac{d}{d x} y = (3 x + 2) (5 x^{2} + 4)

\int \frac{x}{x - 1 + 1} d x

f (x) = 2 e^{- x} x - e^{- x} x^{2}

What is the derivative of (ln(x)^{1/x}) ?
The derivative of (ln(x)^{1/x}) is ln^{1/x}(x)(-(ln(ln(x)))/(x^2)+1/(x^2ln(x)))
What is the first derivative of (ln(x)^{1/x}) ?
The first derivative of (ln(x)^{1/x}) is ln^{1/x}(x)(-(ln(ln(x)))/(x^2)+1/(x^2ln(x)))