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_{0}^{2} \int_{0}^{2 - x^{2}} 18 x y d y d x

x \to - 2 lim (\frac{x + 2}{2 x + x ^{2} + 12})

x \to 3 lim (\frac{( 2 x ^{3} - 5 x ^{2} - 2 x - 3 )}{4 x ^{3} - 13 x ^{2} + 4 x - 3})

x \to 0 lim (x \cdot cos (x))

\int \frac{- 4 x + 4}{5 ( 4 x ^{2} - 4 x + 5 )} d x

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)