What is the derivative of x4^x ?

The derivative of x4^x is 4^x+ln(2)*2^{2x+1}x

What is the first derivative of x4^x ?

The first derivative of x4^x is 4^x+ln(2)*2^{2x+1}x

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\frac{d}{d x} (x 4^{x})

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

Solution steps

\frac{d}{d x} (x \cdot 4^{x})

Apply the Product Rule:

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

f = x, g = 4^{x}

= \frac{d x}{d x} 4^{x} + \frac{d}{d x} (4^{x}) x

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

\frac{d}{d x} (4^{x}) = ln (2) \cdot 2^{2 x + 1}

= 1 \cdot 4^{x} + ln (2) \cdot 2^{2 x + 1} x

Simplify

= 4^{x} + ln (2) \cdot 2^{2 x + 1} x

x^{13}

\frac{d y}{d x}, (7 x y + 3)^{2} = 14 y

y^{^{'}} - y = x y^{2}

\frac{\partial}{\partial x} (\frac{1}{x ^{- 1} + 4 y ^{- 1}})

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

What is the derivative of x4^x ?
The derivative of x4^x is 4^x+ln(2)*2^{2x+1}x
What is the first derivative of x4^x ?
The first derivative of x4^x is 4^x+ln(2)*2^{2x+1}x