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

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

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

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

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

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

Solution steps

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

Simplify

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

\int 2 x^{3} e^{3 x^{4}} d x

\int_{1}^{2} x^{2} + 2 x d x

\int_{0}^{π} sin (5 x) d x

y = (x^{3} - 16 x)^{12}

5 x - 2 x^{2}

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