What is the derivative of (4e^x/(2e^x+1)) ?

The derivative of (4e^x/(2e^x+1)) is (4e^x)/((2e^x+1)^2)

What is the first derivative of (4e^x/(2e^x+1)) ?

The first derivative of (4e^x/(2e^x+1)) is (4e^x)/((2e^x+1)^2)

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Popular >

\frac{d}{d x} (\frac{4 e ^{x}}{2 e ^{x} + 1})

\frac{4 e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 e ^{x}}{2 e ^{x} + 1})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{e ^{x}}{2 e ^{x} + 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 4 \cdot \frac{\frac{d}{d x} ( e ^{x} ) ( 2 e ^{x} + 1 ) - \frac{d}{d x} ( 2 e ^{x} + 1 ) e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

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

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

= 4 \cdot \frac{e ^{x} ( 2 e ^{x} + 1 ) - 2 e ^{x} e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

Simplify

4 \cdot \frac{e ^{x} ( 2 e ^{x} + 1 ) - 2 e ^{x} e ^{x}}{( 2 e ^{x} + 1 ) ^{2}} : \frac{4 e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

= \frac{4 e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

2 sin (2 t)

\frac{d y}{d x} y = \frac{x ^{3}}{3} - x

\frac{\partial}{\partial x} (cos^{6} (x^{8} y^{9}))

y^{^{''}} + 5 y^{^{'}} + 4 y = 0

\int \frac{10}{x ^{2}} d x

What is the derivative of (4e^x/(2e^x+1)) ?
The derivative of (4e^x/(2e^x+1)) is (4e^x)/((2e^x+1)^2)
What is the first derivative of (4e^x/(2e^x+1)) ?
The first derivative of (4e^x/(2e^x+1)) is (4e^x)/((2e^x+1)^2)