What is the derivative of ((x^2-4))/((x^2-1)) ?

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Expand

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

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

\int \frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} d x

n = 0 \sum \infty \frac{n}{n + 2}

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

\int x^{3} x^{2} + 15 d x

n = 1 \sum \infty \frac{1}{2 ^{n} ( n + 1 )}

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