What is (\partial)/(\partial y)(1/(1+e^{-x-2y)}) ?

The answer to (\partial)/(\partial y)(1/(1+e^{-x-2y)}) is (2e^{-2y-x})/((1+e^{-x-2y))^2}

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(\partial)/(\partial y)(1/(1+e^{-x-2y)})

\frac{\partial}{\partial y} (\frac{1}{1 + e ^{- x - 2 y}})

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

Solution steps

\frac{\partial}{\partial y} (\frac{1}{1 + e ^{- x - 2 y}})

Treat

x

as a constant

Apply exponent rule:

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

= \frac{\partial}{\partial y} ((1 + e^{- x - 2 y})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + e ^{- x - 2 y} ) ^{2}} \frac{\partial}{\partial y} (1 + e^{- x - 2 y})

= - \frac{1}{( 1 + e ^{- x - 2 y} ) ^{2}} \frac{\partial}{\partial y} (1 + e^{- x - 2 y})

\frac{\partial}{\partial y} (1 + e^{- x - 2 y}) = - 2 e^{- 2 y - x}

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

Simplify

- \frac{1}{( 1 + e ^{- x - 2 y} ) ^{2}} (- 2 e^{- 2 y - x}) : \frac{2 e ^{- 2 y - x}}{( 1 + e ^{- x - 2 y} ) ^{2}}

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

\frac{d}{d x} (3 (5 - 8 x)^{5})

\frac{d}{d x} (6 x + 3)

x \to 3 lim (2 x^{2} - 4 x + 5)

\frac{d}{d x} (I n (x + 1))

f (x) = 4 π x^{2}

What is (\partial)/(\partial y)(1/(1+e^{-x-2y)}) ?
The answer to (\partial)/(\partial y)(1/(1+e^{-x-2y)}) is (2e^{-2y-x})/((1+e^{-x-2y))^2}