What is the derivative of f(X)=(X^3+3X+2)/(X^2-1) ?

The derivative of f(X)=(X^3+3X+2)/(X^2-1) is (X^4-6X^2-4X-3)/((X^2-1)^2)

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

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

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derivative of

f (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}}

\int_{- 1}^{- 2} \frac{16 - x ^{3}}{x ^{3}} d x

\int 5 sin^{3} (5 x) cos^{3} (5 x) d x

\frac{\partial}{\partial x} (x^{2} y + e^{x y})

\frac{\partial ^{3}}{\partial x \partial y \partial z} (e^{x y z^{9}})

\frac{d}{d t} (sin (t))

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