What is the derivative of 4^x+1 ?

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

What is the first derivative of 4^x+1 ?

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

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\frac{d}{d x} (4^{x} + 1)

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

Solution steps

\frac{d}{d x} (4^{x} + 1)

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (1) = 0

= ln (2) \cdot 2^{2 x + 1} + 0

Simplify

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

\frac{d}{d x} (x^{2} + 3 x + 5)

x \to 2 lim (\frac{4}{( x - 2 ) ^{3}})

\frac{d}{d x} (d cos (x) + x^{2} sin (x))

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{3}}}

∣ x ∣ + 3

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