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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

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