retour
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$f_1'(x) = 1 \times \mathrm e^x + x\times \mathrm e^x = \mathrm e^x + x\mathrm e^x$.
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$f_2'(x) = \dfrac{3(x^2 + 1) - (3x+5)\times (2x)}{(x^2+1)^2}$
$=\dfrac{3x^2 + 3 - 6x^2 - 10x}{(x^2+1)^2}
=\dfrac{-3x^2-10x +3}{(x^2+1)^2}$.
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$f_3'(x) = (2x+1)\mathrm e^{x^2+x+1}$.
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$f_4'(x) = 5 \times 4 \times (4x-3)^4 = 20(4x-3)^{4}$.
retour