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

The derivative of (x^3+3x+2/(x^2-1)) is (x^4-6x^2-4x-3)/((x^2-1)^2)

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

The first derivative of (x^3+3x+2/(x^2-1)) is (x^4-6x^2-4x-3)/((x^2-1)^2)

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

\frac{x ^{4} - 6 x ^{2} - 4 x - 3}{( x ^{2} - 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(3 x^{2} + 3) (x^{2} - 1) - 2 x (x^{3} + 3 x + 2) : x^{4} - 6 x^{2} - 4 x - 3

= \frac{x ^{4} - 6 x ^{2} - 4 x - 3}{( x ^{2} - 1 ) ^{2}}

x^{2} y^{^{''}} - 5 x y^{^{'}} + 13 y = 0

\frac{d}{d x} (sin (cos (x^{2} + 3)))

y = 4 x, y = x^{3} + 3 x^{2}

\int_{0}^{1} \frac{2}{7} x^{2} (x + 3) d x

\int_{0}^{1} π (5 x^{6} - 5 x)^{2} d x

What is the derivative of (x^3+3x+2/(x^2-1)) ?
The derivative of (x^3+3x+2/(x^2-1)) is (x^4-6x^2-4x-3)/((x^2-1)^2)
What is the first derivative of (x^3+3x+2/(x^2-1)) ?
The first derivative of (x^3+3x+2/(x^2-1)) is (x^4-6x^2-4x-3)/((x^2-1)^2)