What is (\partial)/(\partial x)((x^2-2x)/((x-1)^2)) ?

The answer to (\partial)/(\partial x)((x^2-2x)/((x-1)^2)) is 2/((x-1)^3)

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(\partial)/(\partial x)((x^2-2x)/((x-1)^2))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int e^{- b x} d x

\int_{1}^{3} x^{2} ln (x) d x

\int \frac{5 x ^{2} + 2 x - 5}{x ^{3} - x} d x

f (x) = lo g_{10} (1 + x^{2})

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (3 x)

What is (\partial)/(\partial x)((x^2-2x)/((x-1)^2)) ?
The answer to (\partial)/(\partial x)((x^2-2x)/((x-1)^2)) is 2/((x-1)^3)