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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derivative of

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}

1 - x^{2} - ln (x)

\frac{100}{x ^{2}}

ln (1 + x^{\frac{1}{x}})

5 t^{2} - 2 t

\frac{2 x}{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