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

The derivative of f(x)=(2x+x^2)/((1+x)^2) is 2/((1+x)^3)

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

The first derivative of f(x)=(2x+x^2)/((1+x)^2) is 2/((1+x)^3)

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derivative of

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

\frac{2}{( 1 + x ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 + 2 x ) ( 1 + x ) ^{2} - 2 ( 1 + x ) ( 2 x + x ^{2} )}{( ( 1 + x ) ^{2} ) ^{2}} : \frac{2}{( 1 + x ) ^{3}}

= \frac{2}{( 1 + x ) ^{3}}

\frac{\partial}{\partial x} (\frac{Y ( x )}{x})

x \to 0 lim (\frac{3 sin ( 2 x )}{x})

x y^{^{'}} - 4 y = 6 x^{4} e^{x}

f (x) = \frac{- 4 x}{x ^{2} + 1}, at x = - 1

\int_{0}^{1} \frac{2}{x ^{2} + 1} d x

What is the derivative of f(x)=(2x+x^2)/((1+x)^2) ?
The derivative of f(x)=(2x+x^2)/((1+x)^2) is 2/((1+x)^3)
What is the first derivative of f(x)=(2x+x^2)/((1+x)^2) ?
The first derivative of f(x)=(2x+x^2)/((1+x)^2) is 2/((1+x)^3)