What is the derivative of y=(ln(x))/(e^x) ?

The derivative of y=(ln(x))/(e^x) is (1-xln(x))/(e^xx)

What is the first derivative of y=(ln(x))/(e^x) ?

The first derivative of y=(ln(x))/(e^x) is (1-xln(x))/(e^xx)

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derivative of

y = \frac{ln ( x )}{e ^{x}}

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d}{d x} (e^{x}) = e^{x}

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

Simplify

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

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

\int csc (6 x) d x

\frac{2 s}{s ^{2} + 2 s + 1}

\int_{0}^{\frac{π}{4}} (sin (x)) d x

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

f (x) = 8 x (x + 2), at x = 3

What is the derivative of y=(ln(x))/(e^x) ?
The derivative of y=(ln(x))/(e^x) is (1-xln(x))/(e^xx)
What is the first derivative of y=(ln(x))/(e^x) ?
The first derivative of y=(ln(x))/(e^x) is (1-xln(x))/(e^xx)