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

The derivative of f(x)=(x^3)/(x^2-1) is (x^4-3x^2)/((x^2-1)^2)

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int \frac{x}{x ^{2} - 9} d x

\frac{d}{d x} (3^{x^{5} + 2})

\frac{\partial}{\partial x} (2 x sin (x y))

x v^{'} + 2 v = 0

x y^{^{'}} = 2 y + x^{3} cos (x)

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