What is the derivative of 2x^x ?

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

What is the first derivative of 2x^x ?

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

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derivative of

2 x^{x}

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

n = 0 \sum \infty \frac{2 ^{2}}{3 ^{2}}

\frac{\partial}{\partial s} (- 3 s^{2} t^{2})

\frac{d y}{d x} = \frac{y ^{3}}{3 x + 1}, y (0) = - 1

\frac{2}{t ^{3}}

y^{^{'}} + 3 y = x e^{- 2 x}

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