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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (1 \cdot x + 1) = 1

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

Simplify

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

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

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

e^{x} (5 - 4 x + 4 x^{2})

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

y = (7 x + 4) x^{2}

\frac{d}{d t} (s - t)

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