What is the derivative of (2x-3/(x^2-1)) ?

The derivative of (2x-3/(x^2-1)) is (-2x^2+6x-2)/((x^2-1)^2)

What is the first derivative of (2x-3/(x^2-1)) ?

The first derivative of (2x-3/(x^2-1)) is (-2x^2+6x-2)/((x^2-1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x - 3) = 2

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

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

Expand

2 (x^{2} - 1) - 2 x (2 x - 3) : - 2 x^{2} + 6 x - 2

= \frac{- 2 x ^{2} + 6 x - 2}{( x ^{2} - 1 ) ^{2}}

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

\int (6 x + y) d x

x y^{'} + (1 + x cot (x)) y = 0

\int_{- 5}^{0} \frac{1}{4 - x} d x

\int (\frac{1}{x ^{2}} - \frac{4}{x ^{3}}) d x

What is the derivative of (2x-3/(x^2-1)) ?
The derivative of (2x-3/(x^2-1)) is (-2x^2+6x-2)/((x^2-1)^2)
What is the first derivative of (2x-3/(x^2-1)) ?
The first derivative of (2x-3/(x^2-1)) is (-2x^2+6x-2)/((x^2-1)^2)