How do you find the derivative of $- 2 x {(x^{2} + 3)}^{- 2}$ $-2x(x^2+3)^-2$ ?

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Answer 1 · 2016-03-17T16:53:43

$\frac{d}{d x} = \frac{6 x^{2} - 6}{{(x^{2} + 3)}^{3}}$ $d/dx=(6x^2-6)/((x^2+3)^3$

$f = - 2 x, g = {(x^{2} + 3)}^{- 2}$ $f=-2x,g=(x^2+3)^-2$

$f'=-2,g'=-2(x^2+3)^-3 *2x$

$(df)/(dx)=(8x^2)/(x^2+3)^3 -2/((x^2+3)^2$

$(df)/(dx)=(8x^2-2(x^2+3))/((x^2+3)^3$

$(df)/(dx) = (8x^2-2x^2-6)/((x^2+3)^3$

$(df)/(dx)=(6x^2-6)/((x^2+3)^3$

How do you find the derivative of -2x(x^2+3)^-2−2x(x2+3)−2-2x(x^2+3)^-2?