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

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

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

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

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

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

Solution steps

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

Simplify

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

= \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)

(3, - 8), (12, - 15)

\frac{d y}{d x} = - \frac{x}{4 y}

\frac{d}{d x} (x^{2} \cdot y)

\int_{0}^{1} \frac{4}{x ( x + 1 )} d x

y = 4 (x + 1), y = 5 (x + 1), x = 2

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