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

The derivative of y=(e^x)/(x+e^x) is (e^xx-e^x)/((x+e^x)^2)

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Expand

e^{x} (x + e^{x}) - (1 + e^{x}) e^{x} : e^{x} x - e^{x}

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

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

\frac{d}{d x} (x^{3} (x^{2} + 4 + \frac{1}{x}))

\frac{d}{d x} (x^{2} + 2 y^{2} + z^{2} - 6 x + 3 y - 2 z + 5)

y^{^{''}} - 6 y^{^{'}} + 0 y = 0

\int \frac{5}{e ^{x} + 2} d x

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