What is the derivative of x2^x ?

The derivative of x2^x is 2^x+ln(2)*2^xx

What is the first derivative of x2^x ?

The first derivative of x2^x is 2^x+ln(2)*2^xx

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derivative of

x 2^{x}

2^{x} + ln (2) \cdot 2^{x} x

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = 2^{x}

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

\frac{d x}{d x} = 1

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

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

Simplify

= 2^{x} + ln (2) \cdot 2^{x} x

\int sin^{4} (x) \cdot cos^{5} (x) d x

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

f (z) = (9 z^{4} + 3 z^{2} + 1) (3 z^{3} - z)

\int (4 - 2 x)^{2} d x

x \to - 8 lim (9 x + 10)

What is the derivative of x2^x ?
The derivative of x2^x is 2^x+ln(2)*2^xx
What is the first derivative of x2^x ?
The first derivative of x2^x is 2^x+ln(2)*2^xx