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

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

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

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

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

cos (x), at x = \frac{π}{2}

\frac{d}{d r} (r 1 - r^{2})

x \to \infty lim (2 x - 1)

y^{^{'}} = y (1 - y), y (0) = - 2

\int \frac{y + 1}{y ^{2} + y + 1} d y

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