What is the derivative of y=x2^x ?

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

What is the first derivative of y=x2^x ?

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

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derivative of

y = 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

y^{^{''}} + 25 y = cos (5 x)

\frac{d}{d x} ((x - 2) (x^{2} + 4))

\int ln (x^{2} + 13 x + 42) d x

\int \frac{x}{9 x ^{2} + 675} d x

\int \frac{e ^{x}}{x} d x

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