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

The derivative of y=(e^x)/(ln(x)) is (e^xxln(x)-e^x)/(xln^2(x))

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

The first derivative of y=(e^x)/(ln(x)) is (e^xxln(x)-e^x)/(xln^2(x))

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = (8 x^{3} + 2 x^{2})^{4}

\frac{d}{d x} ((1 + 9 x^{2}) (x - x^{2}))

\frac{d}{d x} (arctan (arcsin (x)))

\int 3 y^{5} y^{3} + 1 d y

x \to - 1 lim (x + 2 x^{4})

What is the derivative of y=(e^x)/(ln(x)) ?
The derivative of y=(e^x)/(ln(x)) is (e^xxln(x)-e^x)/(xln^2(x))
What is the first derivative of y=(e^x)/(ln(x)) ?
The first derivative of y=(e^x)/(ln(x)) is (e^xxln(x)-e^x)/(xln^2(x))