What is the derivative of (2x^x) ?

The derivative of (2x^x) is (ln(2x)+1)(2x)^x

What is the first derivative of (2x^x) ?

The first derivative of (2x^x) is (ln(2x)+1)(2x)^x

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

(ln (2 x) + 1) (2 x)^{x}

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

(2 x)^{x} = e^{x l n (2 x)}

= \frac{d}{d x} (e^{x l n (2 x)})

Apply the chain rule:

e^{x l n (2 x)} \frac{d}{d x} (x ln (2 x))

= e^{x l n (2 x)} \frac{d}{d x} (x ln (2 x))

\frac{d}{d x} (x ln (2 x)) = ln (2 x) + 1

= e^{x l n (2 x)} (ln (2 x) + 1)

e^{x l n (2 x)} = (2 x)^{x}

= (ln (2 x) + 1) (2 x)^{x}

y^{^{'''}} - 8 y = 0

\frac{d}{d x} (ln (2 - 5 x))

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

\int sin (7 x) ln (sin (7 x)) d x

What is the derivative of (2x^x) ?
The derivative of (2x^x) is (ln(2x)+1)(2x)^x
What is the first derivative of (2x^x) ?
The first derivative of (2x^x) is (ln(2x)+1)(2x)^x