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

The derivative of 2/(1+3e^{-2x}) is (12e^{-2x})/((1+3e^{-2x))^2}

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

The first derivative of 2/(1+3e^{-2x}) is (12e^{-2x})/((1+3e^{-2x))^2}

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

\frac{12 e ^{- 2 x}}{( 1 + 3 e ^{- 2 x} ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (1 + 3 e^{- 2 x}) = - 6 e^{- 2 x}

= 2 (- \frac{1}{( 1 + 3 e ^{- 2 x} ) ^{2}} (- 6 e^{- 2 x}))

Simplify

2 (- \frac{1}{( 1 + 3 e ^{- 2 x} ) ^{2}} (- 6 e^{- 2 x})) : \frac{12 e ^{- 2 x}}{( 1 + 3 e ^{- 2 x} ) ^{2}}

= \frac{12 e ^{- 2 x}}{( 1 + 3 e ^{- 2 x} ) ^{2}}

y = sin (cos (tan (π x)))

n = 1 \sum \infty \frac{4 ^{n}}{7 ^{n}}

\frac{d}{d x} (0.006 x^{3} + 0.07 x^{2} + 18 x + 600)

\int_{- \infty}^{\infty} \frac{2 x}{( x ^{2} + 1 ) ^{12}} d x

\frac{d y}{d x} = y (2 x - 5)

What is the derivative of 2/(1+3e^{-2x}) ?
The derivative of 2/(1+3e^{-2x}) is (12e^{-2x})/((1+3e^{-2x))^2}
What is the first derivative of 2/(1+3e^{-2x}) ?
The first derivative of 2/(1+3e^{-2x}) is (12e^{-2x})/((1+3e^{-2x))^2}