What is the derivative of 1/(4^x) ?

The derivative of 1/(4^x) is -ln(2)*2^{-2x+1}

What is the first derivative of 1/(4^x) ?

The first derivative of 1/(4^x) is -ln(2)*2^{-2x+1}

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derivative of

\frac{1}{4 ^{x}}

- ln (2) \cdot 2^{- 2 x + 1}

Solution steps

\frac{d}{d x} (\frac{1}{4 ^{x}})

Apply exponent rule:

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

= \frac{d}{d x} (4^{- x})

Apply exponent rule:

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

4^{- x} = e^{(- x) l n (4)}

= \frac{d}{d x} (e^{(- x) l n (4)})

Apply the chain rule:

e^{(- x) l n (4)} \frac{d}{d x} ((- x) ln (4))

= e^{(- x) l n (4)} \frac{d}{d x} ((- x) ln (4))

\frac{d}{d x} ((- x) ln (4)) = - 2 ln (2)

= e^{(- x) l n (4)} (- 2 ln (2))

Simplify

e^{(- x) l n (4)} (- 2 ln (2)) : - ln (2) \cdot 2^{- 2 x + 1}

= - ln (2) \cdot 2^{- 2 x + 1}

\int_{\frac{1}{2}}^{1} \frac{4}{9} x (5 - x^{2}) d x

\int_{1}^{t} 39 x^{- 3} d x

\frac{d}{d x} ((x) arctan (x))

\int_{36}^{x} 1 + 9 x d x

\frac{t - t}{t ^{\frac{1}{7}}}

What is the derivative of 1/(4^x) ?
The derivative of 1/(4^x) is -ln(2)*2^{-2x+1}
What is the first derivative of 1/(4^x) ?
The first derivative of 1/(4^x) is -ln(2)*2^{-2x+1}