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

The derivative of f(x)=(4x+x^2)/((2+x)^2) is 8/((2+x)^3)

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

The first derivative of f(x)=(4x+x^2)/((2+x)^2) is 8/((2+x)^3)

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derivative of

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

\frac{8}{( 2 + x ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 4 + 2 x ) ( 2 + x ) ^{2} - 2 ( 2 + x ) ( 4 x + x ^{2} )}{( ( 2 + x ) ^{2} ) ^{2}} : \frac{8}{( 2 + x ) ^{3}}

= \frac{8}{( 2 + x ) ^{3}}

\int_{0}^{3} 17 d x

6 y^{^{''}} + 6 y^{^{'}} - 12 y = 0

y = 2 x^{3} + 3 x^{2} - 12 x + 7

\int \frac{12 x}{1 + 4 x} d x

\frac{d}{d x} (\frac{x ^{3}}{3} + x^{2} + 4 x - 10)

What is the derivative of f(x)=(4x+x^2)/((2+x)^2) ?
The derivative of f(x)=(4x+x^2)/((2+x)^2) is 8/((2+x)^3)
What is the first derivative of f(x)=(4x+x^2)/((2+x)^2) ?
The first derivative of f(x)=(4x+x^2)/((2+x)^2) is 8/((2+x)^3)