What is the derivative of 2^x+1 ?

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

What is the first derivative of 2^x+1 ?

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

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\frac{d}{d x} (2^{x} + 1)

ln (2) \cdot 2^{x}

Solution steps

\frac{d}{d x} (2^{x} + 1)

Apply the Sum/Difference Rule:

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

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

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

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

= 2^{x} ln (2) + 0

Simplify

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

\frac{d y}{d t} = - \frac{y}{t ^{2}}

f (x) = 6 x - x^{2}, at x = 4

\frac{d}{d x} ((x - 5) sin (x))

n = 0 \sum \infty \frac{1}{1. 5 ^{n}}

\int - x^{2} - 2 x d x

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