What is the derivative of y=x^{1/x} ?

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

What is the first derivative of y=x^{1/x} ?

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

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derivative of

y = x^{\frac{1}{x}}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int e^{- 4 x} sin (x) d x

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} + y ^{2} + z ^{2}})

\int_{0}^{4} x e^{- 0.1 x} d x

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

x \to + 0 lim (\frac{( 1 + x ) ^{4} - 1}{x})

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