What is the derivative of ln(xe^x) ?

The derivative of ln(xe^x) is (e^x)/x+e^xln(x)

What is the first derivative of ln(xe^x) ?

The first derivative of ln(xe^x) is (e^x)/x+e^xln(x)

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

\frac{e ^{x}}{x} + e^{x} ln (x)

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = ln (x), g = e^{x}

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

\frac{d}{d x} (ln (x)) = \frac{1}{x}

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

= \frac{1}{x} e^{x} + e^{x} ln (x)

\frac{1}{x} e^{x} = \frac{e ^{x}}{x}

= \frac{e ^{x}}{x} + e^{x} ln (x)

\frac{d}{d x} ((5 x - 4)^{2})

f (x) = ln (\frac{1}{x})

\frac{d y}{d t} = y - 2 t

\int \frac{x}{x + 8} d x

x \to 2 lim (\frac{1}{2 - 5 x})

What is the derivative of ln(xe^x) ?
The derivative of ln(xe^x) is (e^x)/x+e^xln(x)
What is the first derivative of ln(xe^x) ?
The first derivative of ln(xe^x) is (e^x)/x+e^xln(x)