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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{x}{3 x + 8} d x

e^{3 t} cos (6 t)

\int 3 a^{7} x^{6} d x

x \to 1 lim (e^{(x^{2} - x)})

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