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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derivative of

(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}

\int \frac{2}{16 + 3 x ^{2}} d x

re g a r d s p \frac{I}{2 p}

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\frac{d ^{2} y}{d x ^{2}} = 4 x

\int \frac{6 x ^{2} + x + 6}{( x ^{2} + 1 ) ^{2}} 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