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Popular >

derivative of (1+e^{-t})^{-1}+e^{-1}

Solution

derivative of

(1 + e^{- t})^{- 1} + e^{- 1}

Solution

\frac{e ^{- t}}{( e ^{- t} + 1 ) ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} ((1 + e^{- t})^{- 1}) = \frac{e ^{- t}}{( 1 + e ^{- t} ) ^{2}}

\frac{d}{d t} (e^{- 1}) = 0

= \frac{e ^{- t}}{( 1 + e ^{- t} ) ^{2}} + 0

Simplify

= \frac{e ^{- t}}{( e ^{- t} + 1 ) ^{2}}

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\frac{d y}{d x} y = \frac{sec ( x )}{1 + sec ( x )}

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Frequently Asked Questions (FAQ)

What is the derivative of (1+e^{-t})^{-1}+e^{-1} ?
The derivative of (1+e^{-t})^{-1}+e^{-1} is (e^{-t})/((e^{-t)+1)^2}
What is the first derivative of (1+e^{-t})^{-1}+e^{-1} ?
The first derivative of (1+e^{-t})^{-1}+e^{-1} is (e^{-t})/((e^{-t)+1)^2}