What is the derivative of 2^{-3t} ?

The derivative of 2^{-3t} is -3ln(2)*8^{-t}

What is the first derivative of 2^{-3t} ?

The first derivative of 2^{-3t} is -3ln(2)*8^{-t}

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derivative of

2^{- 3 t}

- 3 ln (2) \cdot 8^{- t}

Solution steps

\frac{d}{d t} (2^{- 3 t})

Simplify

2^{- 3 t} : 8^{- t}

= \frac{d}{d t} (8^{- t})

Apply exponent rule:

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

8^{- t} = e^{(- t) l n (8)}

= \frac{d}{d t} (e^{(- t) l n (8)})

Apply the chain rule:

e^{(- t) l n (8)} \frac{d}{d t} ((- t) ln (8))

= e^{(- t) l n (8)} \frac{d}{d t} ((- t) ln (8))

\frac{d}{d t} ((- t) ln (8)) = - 3 ln (2)

= e^{(- t) l n (8)} (- 3 ln (2))

Simplify

e^{(- t) l n (8)} (- 3 ln (2)) : - 3 ln (2) \cdot 8^{- t}

= - 3 ln (2) \cdot 8^{- t}

x^{x} (lo g_{10} (x) + 1)

f (x) = x^{5} + x^{2} - 2 x + 1

\frac{d}{d t} (2 t (\frac{n}{4}) + c)

ln (tan (\frac{x}{2}))

4 e^{2} x

What is the derivative of 2^{-3t} ?
The derivative of 2^{-3t} is -3ln(2)*8^{-t}
What is the first derivative of 2^{-3t} ?
The first derivative of 2^{-3t} is -3ln(2)*8^{-t}