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

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

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

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

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derivative of

\frac{x ^{2} + 1}{x ^{2} - 1}

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

Solution steps

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

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

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

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

Simplify

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

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

y = x^{2} + 5, (4, 21)

u \to 0 lim (ln (\frac{u + 1}{u}))

y = - x^{3} + 2 x^{2} - 4, at x = - 1

(- 4.3) (5.3)

n = 1 \sum \infty \frac{5}{3 ^{n} + 1}

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