What is the derivative of 2^{-2x} ?

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

What is the first derivative of 2^{-2x} ?

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

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

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

Solution steps

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

Simplify

2^{- 2 x} : 4^{- x}

= \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}

f (x) = cos (x) - sin (x)

n = 0 \sum \infty \frac{2 n}{n} x^{n}

\frac{\partial}{\partial x} (e^{x} sin (y + z) - z)

x \to 4 lim (3 - 2)

f (x) = 4 - x

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