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 \frac{14 - 14 x}{1 - x} d x

\frac{d y}{d t} = - \frac{t}{e ^{y + t^{2}}}

\int \frac{3 + x + x}{x} d x

\frac{d}{d x} (arccosh (x))

θ \to 0 lim (\frac{sin ( 3 θ )}{2 θ})

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)