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

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

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

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

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derivative of

y = \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}}

\frac{d}{d t} (10 + 5 t + 2 t^{2})

\frac{d}{d n} ((1 - i)^{n - 1})

(\frac{x}{x + 1}) cos (x)

4 - 3 u^{2}

60 t^{2} - 60 t

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