What is the ((e^x)/x)^' ?

The ((e^x)/x)^' is (e^xx-e^x)/(x^2)

What is the first ((e^x)/x)^' ?

The first ((e^x)/x)^' is (e^xx-e^x)/(x^2)

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(\frac{e ^{x}}{x})^{^{'}}

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

Solution steps

(\frac{e ^{x}}{x})^{'}

Apply the Quotient Rule:

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

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

(e^{x})^{'} = e^{x}

x^{'} = 1

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

Simplify

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

f (t) = (3 t + 14)^{\frac{2}{3}}

\frac{d y}{d x} + \frac{y}{x ^{2} + x} = x + 1

\frac{d}{d x} (2 I n x)

\frac{d}{d x} (3 e^{- 4 x})

\frac{1}{x}, 2