What is the derivative of 1-2^x ?

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

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

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

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derivative of

1 - 2^{x}

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

y = x^{2} (2 x + 3)^{5}

\int \frac{e ^{2 x}}{25 + e ^{4 x}} d x

y = x^{2} 4 x - 7

f (x) = 6 sec (x), at x = \frac{π}{6}

n = 1 \sum \infty n e^{4 n}

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