What is the derivative of (e^x/(6+e^x)) ?

The derivative of (e^x/(6+e^x)) is (6e^x)/((6+e^x)^2)

What is the first derivative of (e^x/(6+e^x)) ?

The first derivative of (e^x/(6+e^x)) is (6e^x)/((6+e^x)^2)

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\frac{d}{d x} (\frac{e ^{x}}{6 + e ^{x}})

\frac{6 e ^{x}}{( 6 + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{6 + 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} ) ( 6 + e ^{x} ) - \frac{d}{d x} ( 6 + e ^{x} ) e ^{x}}{( 6 + e ^{x} ) ^{2}}

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

\frac{d}{d x} (6 + e^{x}) = e^{x}

= \frac{e ^{x} ( 6 + e ^{x} ) - e ^{x} e ^{x}}{( 6 + e ^{x} ) ^{2}}

Simplify

\frac{e ^{x} ( 6 + e ^{x} ) - e ^{x} e ^{x}}{( 6 + e ^{x} ) ^{2}} : \frac{6 e ^{x}}{( 6 + e ^{x} ) ^{2}}

= \frac{6 e ^{x}}{( 6 + e ^{x} ) ^{2}}

\frac{d}{d x} (\frac{64 x}{x ^{2} + 64})

\int 4 (1 + tan^{2} (x)) d x

y^{^{''}} + 2 y^{^{'}} = 2 x + 5 - e^{-} 2 x

\frac{d}{d x} (\frac{2 x ^{3} - 7}{x ^{2}})

\frac{d y}{d x} = 100 - y

What is the derivative of (e^x/(6+e^x)) ?
The derivative of (e^x/(6+e^x)) is (6e^x)/((6+e^x)^2)
What is the first derivative of (e^x/(6+e^x)) ?
The first derivative of (e^x/(6+e^x)) is (6e^x)/((6+e^x)^2)