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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{x - 1}) + \frac{d}{d x} (\frac{1}{x + 1})

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

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

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

Simplify

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

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

\int \frac{1}{x - 5} d x

\frac{d}{d x} (x + π)

y = \frac{1}{x ^{9}}

\int \frac{1}{( 8 x - 1 ) ^{2}} d x

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

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