What is the derivative of f(x)=(2x)/((x^2-x)) ?

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

What is the first derivative of f(x)=(2x)/((x^2-x)) ?

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{d}{d x} (\frac{1}{3 x})

\int 7^{x} - sec^{2} (x) d x

\int sin (\frac{π x}{a}) d x

\frac{d}{d x} (\frac{x ^{2} - 6 x - 1}{( x - 3 ) ^{2}})

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = 1

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