What is the derivative of (4e^x/(ln(x))) ?

The derivative of (4e^x/(ln(x))) is (4(e^xxln(x)-e^x))/(xln^2(x))

What is the first derivative of (4e^x/(ln(x))) ?

The first derivative of (4e^x/(ln(x))) is (4(e^xxln(x)-e^x))/(xln^2(x))

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\frac{d}{d x} (\frac{4 e ^{x}}{ln ( x )})

\frac{4 ( e ^{x} x ln ( x ) - e ^{x} )}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{4 e ^{x}}{ln ( x )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \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}}

= 4 \cdot \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}

= 4 \cdot \frac{e ^{x} ln ( x ) - \frac{1}{x} e ^{x}}{( ln ( x ) ) ^{2}}

Simplify

4 \cdot \frac{e ^{x} ln ( x ) - \frac{1}{x} e ^{x}}{ln ^{2} ( x )} : \frac{4 ( e ^{x} x ln ( x ) - e ^{x} )}{x ln ^{2} ( x )}

= \frac{4 ( e ^{x} x ln ( x ) - e ^{x} )}{x ln ^{2} ( x )}

\int y^{2} cos (x) - 3 x^{2} y - 2 x d x

\frac{\partial}{\partial x} (e^{- 7 x} cos (10 x))

f (x) = 10 - 4 e^{x}, (0, 6)

\frac{d}{d x} (sin (4))

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

What is the derivative of (4e^x/(ln(x))) ?
The derivative of (4e^x/(ln(x))) is (4(e^xxln(x)-e^x))/(xln^2(x))
What is the first derivative of (4e^x/(ln(x))) ?
The first derivative of (4e^x/(ln(x))) is (4(e^xxln(x)-e^x))/(xln^2(x))