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

The derivative of x-ln(xe^x) is -1/x

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

The first derivative of x-ln(xe^x) is -1/x

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

- \frac{1}{x}

Solution steps

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

Simplify

x - ln (x e^{x}) : - ln (x)

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

Take the constant out:

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

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

Apply the common derivative:

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

= - \frac{1}{x}

\frac{d}{d x} (2 - \frac{2}{x})

\int \frac{1}{x ^{\frac{2}{5}}} d x

\int_{0}^{2} x^{2} d x

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

y^{^{'}} = e^{5 y + 8 x}, y (0) = 3

What is the derivative of x-ln(xe^x) ?
The derivative of x-ln(xe^x) is -1/x
What is the first derivative of x-ln(xe^x) ?
The first derivative of x-ln(xe^x) is -1/x