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

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

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

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

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\frac{d}{d x} (e^{(\frac{1}{x}) (l n (1 - 2 x))})

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

Solution steps

\frac{d}{d x} (e^{(\frac{1}{x}) (l n (1 - 2 x))})

Simplify

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

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\int \frac{10}{4 x ^{3} - 4 x ^{2} + 5 x} d x

\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

\frac{\partial}{\partial y} (ln (x y))

\int x^{2} 3 - 2 x d x

\int \frac{1}{( a ^{2} + v ^{2} )} d v

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