What is the derivative of (e^x/(6x)) ?

The derivative of (e^x/(6x)) is (e^xx-e^x)/(6x^2)

What is the first derivative of (e^x/(6x)) ?

The first derivative of (e^x/(6x)) is (e^xx-e^x)/(6x^2)

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\frac{d}{d x} (\frac{e ^{x}}{6 x})

\frac{e ^{x} x - e ^{x}}{6 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{6 x})

Take the constant out:

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

= \frac{1}{6} \frac{d}{d x} (\frac{e ^{x}}{x})

Apply the Quotient Rule:

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

= \frac{1}{6} \cdot \frac{\frac{d}{d x} ( e ^{x} ) x - \frac{d x}{d x} e ^{x}}{x ^{2}}

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

\frac{d x}{d x} = 1

= \frac{1}{6} \cdot \frac{e ^{x} x - 1 \cdot e ^{x}}{x ^{2}}

Simplify

\frac{1}{6} \cdot \frac{e ^{x} x - 1 \cdot e ^{x}}{x ^{2}} : \frac{e ^{x} x - e ^{x}}{6 x ^{2}}

= \frac{e ^{x} x - e ^{x}}{6 x ^{2}}

\frac{d}{d x} (x y) = \frac{d}{d x} (4)

f (x) = (arccos (x))^{3}

\int cos (3 x + 2) d x

f (x) = 4 (x - \frac{1}{x})^{3}, at x = 2

y = \frac{- 2 x}{x ^{2} + 1}, (0, 0)

What is the derivative of (e^x/(6x)) ?
The derivative of (e^x/(6x)) is (e^xx-e^x)/(6x^2)
What is the first derivative of (e^x/(6x)) ?
The first derivative of (e^x/(6x)) is (e^xx-e^x)/(6x^2)