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}

\int 48 sec^{4} (x) d x

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

\frac{d}{d q} (- 0.1 q^{2} + 1.2 q + 46.4)

n = 1 \sum \infty (\frac{9}{8})^{n}

\frac{d y}{d x}, x y + 3 = y

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