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

The derivative of (x^2-2x/(x+1)) is (x^2+2x-2)/((x+1)^2)

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

The first derivative of (x^2-2x/(x+1)) is (x^2+2x-2)/((x+1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{'}} = - \frac{y}{x} + e^{x}

f (x) = sin (3 x) - cos (2 x)

\frac{\partial}{\partial x} (x^{3} + 4)

\int_{0}^{1} e^{- s t} t d t

y^{^{'}} = - 2 x tan (y)

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