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

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

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

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

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derivative of

y = \frac{3}{e ^{x} + e ^{- x}}

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (3 x, - 2 y, x - y)

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

y^{^{''}} + y = - sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

\int \frac{ln ( x ^{20} )}{x} d x

\int \frac{1}{( x ^{2} + 1 ) ( x + 1 )} d x

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