Réponses

2013-12-22T17:08:16+01:00
Bonjour,

1) Tu appliques la formule  (\dfrac{1}{u})'=\dfrac{-u'}{u^2}

(\dfrac{1}{x}-\dfrac{1}{x^2})'=(\dfrac{1}{x})'-(\dfrac{1}{x^2})'\\\\(\dfrac{1}{x}-\dfrac{1}{x^2})'=\dfrac{-x'}{x^2}-\dfrac{(-x^2)'}{(x^2)^2}\\\\(\dfrac{1}{x}-\dfrac{1}{x^2})'=\dfrac{-1}{x^2}-\dfrac{-2x}{x^4}\\\\(\dfrac{1}{x}-\dfrac{1}{x^2})'=\dfrac{-1}{x^2}+\dfrac{2x}{x^4}\\\\(\dfrac{1}{x}-\dfrac{1}{x^2})'=\dfrac{-1}{x^2}+\dfrac{2}{x^3}\\\\(\dfrac{1}{x}-\dfrac{1}{x^2})'=\dfrac{-x+2}{x^3}

2) Tu appliques la formule  (\dfrac{u}{v})'=\dfrac{u'v-v'u}{v^2}

(\dfrac{x+3}{x^2+1})'=\dfrac{(x+3)'\times(x^2+1)-(x^2+1)'\times (x+3)}{(x^2+1)^2}\\\\(\dfrac{x+3}{x^2+1})'=\dfrac{1\times(x^2+1)-2x\times (x+3)}{(x^2+1)^2}\\\\(\dfrac{x+3}{x^2+1})'=\dfrac{x^2+1-2x^2-6x}{(x^2+1)^2}\\\\(\dfrac{x+3}{x^2+1})'=\dfrac{-x^2-6x+1}{(x^2+1)^2}