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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int \frac{4 x ^{6} + 3 x ^{5} - 8}{x ^{5}} d x

\frac{d y}{d x} = \frac{( e ^{x} - 1 )}{e ^{y}}

\frac{\partial}{\partial y} (1 + ln (x) + \frac{y}{x})

\int e^{4 x^{2}} d x

\frac{d x}{d y} = \frac{x}{y}

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