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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{- x} = e^{(- x) l n (2)}

= \frac{d}{d x} (e^{(- x) l n (2)})

Apply the chain rule:

e^{(- x) l n (2)} \frac{d}{d x} ((- x) ln (2))

= e^{(- x) l n (2)} \frac{d}{d x} ((- x) ln (2))

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

= e^{(- x) l n (2)} (- ln (2))

Simplify

e^{(- x) l n (2)} (- ln (2)) : - ln (2) \cdot 2^{- x}

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

\int_{0}^{\frac{π}{27}} x tan^{2} (9 x) d x

x \to 0 lim (\frac{tan ( 10 x )}{x})

\frac{d}{d x} (sin (\frac{1}{x - 1}))

y^{^{''}} + 4 y = 3 sin (2 t)

5^{x^{5}}

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}