What is the derivative of x^4+4^x ?

The derivative of x^4+4^x is 4x^3+ln(2)*2^{2x+1}

What is the first derivative of x^4+4^x ?

The first derivative of x^4+4^x is 4x^3+ln(2)*2^{2x+1}

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

4 x^{3} + ln (2) \cdot 2^{2 x + 1}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

= 4 x^{3} + ln (2) \cdot 2^{2 x + 1}

x \to + 0 lim (\frac{e ^{- 5 x} - 1}{x})

\frac{\partial}{\partial x} (e^{x y} + x^{2} z)

\int cos^{2} (θ) d θ

\frac{d ^{2}}{d x ^{2}} f (x) = \frac{7}{x + 4}, at x = 0

n = 1 \sum \infty \frac{n ^{2}}{n !}

What is the derivative of x^4+4^x ?
The derivative of x^4+4^x is 4x^3+ln(2)*2^{2x+1}
What is the first derivative of x^4+4^x ?
The first derivative of x^4+4^x is 4x^3+ln(2)*2^{2x+1}