What is (d^3)/(dx^3)(1/(x^2+1)) ?

The answer to (d^3)/(dx^3)(1/(x^2+1)) is (24x(-x^2+1))/((x^2+1)^4)

(d^3)/(dx^3)(1/(x^2+1))

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x ^{2} + 1})

\frac{24 x ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{4}}

Solution steps

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x ^{2} + 1})

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

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

\frac{d}{d x} (- \frac{2 x}{( x ^{2} + 1 ) ^{2}}) = - \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}

= \frac{d}{d x} (- \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}})

\frac{d}{d x} (- \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}) = \frac{24 x ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{4}}

= \frac{24 x ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{4}}

t an g e n t 8 x^{2} + 3 x - 7

\int_{- 1}^{1} (x + 1) (x + 2) (x + 3) d x

d er i v a t i v e 8 (e^{0} cos (0) - e^{0} sin (0))

t an g e n t \frac{x ^{2}}{x + 9}

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What is (d^3)/(dx^3)(1/(x^2+1)) ?
The answer to (d^3)/(dx^3)(1/(x^2+1)) is (24x(-x^2+1))/((x^2+1)^4)