What is the derivative of (10/(1+e^{-3x)}) ?

The derivative of (10/(1+e^{-3x)}) is (30e^{-3x})/((1+e^{-3x))^2}

What is the first derivative of (10/(1+e^{-3x)}) ?

The first derivative of (10/(1+e^{-3x)}) is (30e^{-3x})/((1+e^{-3x))^2}

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\frac{d}{d x} (\frac{10}{1 + e ^{- 3 x}})

\frac{30 e ^{- 3 x}}{( 1 + e ^{- 3 x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{10}{1 + e ^{- 3 x}})

Take the constant out:

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

= 10 \frac{d}{d x} (\frac{1}{1 + e ^{- 3 x}})

Apply exponent rule:

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

= 10 \frac{d}{d x} ((1 + e^{- 3 x})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + e ^{- 3 x} ) ^{2}} \frac{d}{d x} (1 + e^{- 3 x})

= - \frac{1}{( 1 + e ^{- 3 x} ) ^{2}} \frac{d}{d x} (1 + e^{- 3 x})

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

= 10 (- \frac{1}{( 1 + e ^{- 3 x} ) ^{2}} (- 3 e^{- 3 x}))

Simplify

10 (- \frac{1}{( 1 + e ^{- 3 x} ) ^{2}} (- 3 e^{- 3 x})) : \frac{30 e ^{- 3 x}}{( 1 + e ^{- 3 x} ) ^{2}}

= \frac{30 e ^{- 3 x}}{( 1 + e ^{- 3 x} ) ^{2}}

\frac{d}{d x} (e^{2 x} \cdot 4)

y = x, y = \frac{1}{2} x, x = 25

sin (6 (t - \frac{π}{2}))

\int (3 x + 5)^{2} (\frac{7 x - 4}{x ^{2}}) d x

x \to + 0 lim (- \frac{7}{1 - 7 x})

What is the derivative of (10/(1+e^{-3x)}) ?
The derivative of (10/(1+e^{-3x)}) is (30e^{-3x})/((1+e^{-3x))^2}
What is the first derivative of (10/(1+e^{-3x)}) ?
The first derivative of (10/(1+e^{-3x)}) is (30e^{-3x})/((1+e^{-3x))^2}