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

The derivative of (2x-1)/(2sqrt((x^2-x))) is -1/(4(x^2-x)sqrt(x^2-x))

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

The first derivative of (2x-1)/(2sqrt((x^2-x))) is -1/(4(x^2-x)sqrt(x^2-x))

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derivative of

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

- \frac{1}{4 ( x ^{2} - x ) x ^{2} - x}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{1}{2} \cdot \frac{2 x ^{2} - x - \frac{2 x - 1}{2 x ^{2} - x} ( 2 x - 1 )}{( x ^{2} - x ) ^{2}} : - \frac{1}{4 ( x ^{2} - x ) x ^{2} - x}

= - \frac{1}{4 ( x ^{2} - x ) x ^{2} - x}

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

\int - 2 sin (n x) d x

4 x^{2} = ln (x - y) + 3, at x = 0

\int e^{x} (4 + e^{x})^{5} d x

(x^{''} + x)^{^{'}}

What is the derivative of (2x-1)/(2sqrt((x^2-x))) ?
The derivative of (2x-1)/(2sqrt((x^2-x))) is -1/(4(x^2-x)sqrt(x^2-x))
What is the first derivative of (2x-1)/(2sqrt((x^2-x))) ?
The first derivative of (2x-1)/(2sqrt((x^2-x))) is -1/(4(x^2-x)sqrt(x^2-x))