What is the derivative of y= 1/((1+c_{1)e^{-x})} ?

The derivative of y= 1/((1+c_{1)e^{-x})} is (c_{1}e^{-x})/((1+c_{1)e^{-x})^2}

What is the first derivative of y= 1/((1+c_{1)e^{-x})} ?

The first derivative of y= 1/((1+c_{1)e^{-x})} is (c_{1}e^{-x})/((1+c_{1)e^{-x})^2}

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derivative of

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

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

Solution steps

\frac{d}{d x} (\frac{1}{1 + c _{1} e ^{- x}})

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{( 1 + c _{1} e ^{- x} ) ^{2}} \frac{d}{d x} (1 + c_{1} e^{- x})

= - \frac{1}{( 1 + c _{1} e ^{- x} ) ^{2}} \frac{d}{d x} (1 + c_{1} e^{- x})

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

= - \frac{1}{( 1 + c _{1} e ^{- x} ) ^{2}} (- c_{1} e^{- x})

Simplify

- \frac{1}{( 1 + c _{1} e ^{- x} ) ^{2}} (- c_{1} e^{- x}) : \frac{c _{1} e ^{- x}}{( 1 + c _{1} e ^{- x} ) ^{2}}

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

\frac{\partial}{\partial x} (e^{x t})

\frac{d}{d x} (\frac{1}{x ^{\frac{1}{3}}})

\int z^{2} sin (x) d x

y^{^{''''}} + y = 0

\frac{d y}{d x} = \frac{y ^{2}}{2 x + 3}

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