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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = - 2 sin (2 x)

\int 6 sec^{2} (x) d x

\int x^{3} (ln (x))^{2} d x

\frac{s}{( s ^{2} + 4 )}

x \to 1 lim ((x^{2} - 2)^{3})

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