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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

5 t^{2} - 2 t

\frac{2 x}{y}

\frac{- 2}{t ^{2}}

\frac{d}{d n} (sin (n) cos (n) + 3 n + 1)

(s^{2} + a)^{- 2}

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