What is the derivative of (e^x-1/(2x)) ?

The derivative of (e^x-1/(2x)) is (e^xx-e^x+1)/(2x^2)

What is the first derivative of (e^x-1/(2x)) ?

The first derivative of (e^x-1/(2x)) is (e^xx-e^x+1)/(2x^2)

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

\frac{e ^{x} x - e ^{x} + 1}{2 x ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{e ^{x} x - e ^{x} + 1}{2 x ^{2}}

\int \frac{x}{e ^{7} x} d x

\int_{0}^{\frac{π}{4}} 6 tan^{4} (x) d x

x \cdot arcsin (x)

f (x) = x e^{2 x} sin (6 x)

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

What is the derivative of (e^x-1/(2x)) ?
The derivative of (e^x-1/(2x)) is (e^xx-e^x+1)/(2x^2)
What is the first derivative of (e^x-1/(2x)) ?
The first derivative of (e^x-1/(2x)) is (e^xx-e^x+1)/(2x^2)