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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

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

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

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

y^{^{'}} = 0.3 y (1 - \frac{y}{2500}) - 120

\frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4}

x y^{^{'}} + 4 y = 8 x^{4}

\int e^{- t} t d t

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