What is the derivative of 1/(3(1+e^{-t))^3} ?

The derivative of 1/(3(1+e^{-t))^3} is (e^{-t})/((1+e^{-t))^4}

What is the first derivative of 1/(3(1+e^{-t))^3} ?

The first derivative of 1/(3(1+e^{-t))^3} is (e^{-t})/((1+e^{-t))^4}

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derivative of

\frac{1}{3 ( 1 + e ^{- t} ) ^{3}}

\frac{e ^{- t}}{( 1 + e ^{- t} ) ^{4}}

Solution steps

\frac{d}{d t} (\frac{1}{3 ( 1 + e ^{- t} ) ^{3}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{3} \frac{d}{d t} (\frac{1}{( 1 + e ^{- t} ) ^{3}})

Apply exponent rule:

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

= \frac{1}{3} \frac{d}{d t} ((1 + e^{- t})^{- 3})

Apply the chain rule:

- \frac{3}{( 1 + e ^{- t} ) ^{4}} \frac{d}{d t} (1 + e^{- t})

= - \frac{3}{( 1 + e ^{- t} ) ^{4}} \frac{d}{d t} (1 + e^{- t})

\frac{d}{d t} (1 + e^{- t}) = - e^{- t}

= \frac{1}{3} (- \frac{3}{( 1 + e ^{- t} ) ^{4}} (- e^{- t}))

Simplify

\frac{1}{3} (- \frac{3}{( 1 + e ^{- t} ) ^{4}} (- e^{- t})) : \frac{e ^{- t}}{( 1 + e ^{- t} ) ^{4}}

= \frac{e ^{- t}}{( 1 + e ^{- t} ) ^{4}}

x d y + 3 y d x = x y d y

x \to + 0 lim (\frac{- 6}{x ^{7}})

\int_{0}^{3} ∣ x - 2 ∣ d x

\int \frac{1}{x ^{2} - 2 x + 65} d x

What is the derivative of 1/(3(1+e^{-t))^3} ?
The derivative of 1/(3(1+e^{-t))^3} is (e^{-t})/((1+e^{-t))^4}
What is the first derivative of 1/(3(1+e^{-t))^3} ?
The first derivative of 1/(3(1+e^{-t))^3} is (e^{-t})/((1+e^{-t))^4}