What is the derivative of log_{e}(x^x) ?

The derivative of log_{e}(x^x) is ln(x)+1

What is the first derivative of log_{e}(x^x) ?

The first derivative of log_{e}(x^x) is ln(x)+1

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\frac{d}{d x} (lo g_{e} (x^{x}))

ln (x) + 1

Solution steps

\frac{d}{d x} (lo g_{e} (x^{x}))

Simplify

lo g_{e} (x^{x}) : ln (x^{x})

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{x ^{x}} x^{x} (ln (x) + 1) : ln (x) + 1

= ln (x) + 1

cos^{2} (θ)

\frac{d y}{d x} = y (950 - y)

\int \frac{7}{x + 7 + x} d x

\frac{d}{d x} (\frac{x ^{3} y - x y ^{3}}{x ^{2} + y ^{2}})

\int \frac{8 x}{x ^{2} + 4} d x

What is the derivative of log_{e}(x^x) ?
The derivative of log_{e}(x^x) is ln(x)+1
What is the first derivative of log_{e}(x^x) ?
The first derivative of log_{e}(x^x) is ln(x)+1