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}} \cdot 2 x

Simplify

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

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

\int_{- 1}^{5} 3 x^{2} + 6 x d x

y = 4 x^{3} - 4 x, (- 1, 0)

x \to - 2 lim (- 3)

\int_{0}^{1} (\frac{1}{x ^{2}}) d x

\int cot^{3} (\frac{x}{2}) csc^{2} (\frac{x}{2}) d x