What is the derivative of (2x/(4+x)) ?

The derivative of (2x/(4+x)) is 8/((4+x)^2)

What is the first derivative of (2x/(4+x)) ?

The first derivative of (2x/(4+x)) is 8/((4+x)^2)

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

\frac{8}{( 4 + x ) ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

\frac{d}{d x} (4 + x) = 1

= 2 \cdot \frac{1 \cdot ( 4 + x ) - 1 \cdot x}{( 4 + x ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( 4 + x ) - 1 \cdot x}{( 4 + x ) ^{2}} : \frac{8}{( 4 + x ) ^{2}}

= \frac{8}{( 4 + x ) ^{2}}

\int \frac{1}{4 + t ^{2}} d t

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

f (x) = \frac{6 x + 14}{x + 7}

\int \frac{5}{sec ( x ) - 1} d x

\int \frac{1}{4 x - x ^{2}} d x

What is the derivative of (2x/(4+x)) ?
The derivative of (2x/(4+x)) is 8/((4+x)^2)
What is the first derivative of (2x/(4+x)) ?
The first derivative of (2x/(4+x)) is 8/((4+x)^2)