What is the derivative of f(x)=(e^x)/(1+e^x) ?

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

What is the first derivative of f(x)=(e^x)/(1+e^x) ?

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{x ^{2}}{( 16 - x ^{3} ) ^{2}} d x

x \to - 4 + lim (\frac{x + 3}{x + 4})

\int x^{2} sec^{2} (x) tan (x) d x

y = x^{2} - 4, y = 2 - 5 x^{2}

x \to 0 lim (\frac{6 sin ( 2 x )}{x})

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