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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Expand

(2 x + 1) (2 x - 1) - 2 (x^{2} + x) : 2 x^{2} - 2 x - 1

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

(3 x^{2} - 1) (2 x + 3)

(2^{x})^{x}

(e^{x})^{- 2}

\frac{( e ^{x} + 1 )}{( e ^{x} - 1 )}

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

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