What is the (1/(x^2-1))^' ?

The (1/(x^2-1))^' is -(2x)/((x^2-1)^2)

What is the first (1/(x^2-1))^' ?

The first (1/(x^2-1))^' is -(2x)/((x^2-1)^2)

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= ((x^{2} - 1)^{- 1})^{^{'}}

Apply the chain rule:

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

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

(x^{2} - 1)^{'} = 2 x

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

Simplify

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

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

f (x) = 4 - 3 x^{2}, (5, - 71)

\frac{- 0.5}{s ^{2} + s + 1}

\int x - 2 + \frac{3}{( 3 x - 5 ) ^{6}} d x

\frac{t}{t ^{2} + 8}

(2 x - 7) 4 (x^{2} + x + 1) 5