salut,
1) rappel : (u\v)' = (u'v-v'u) \ v²
donc : f'(x) =[ (x³ + 2x²) \ (x²-1) ]'
= [(3x²+4x) (x²-1) - 2x (x³+2x²)] \ (x²-1)²
= (3[tex] x^{4} [/tex] -3x²+4x³-4x-2[tex] x^{4} [/tex] -4x³) \ (x²-1)²
= ([tex] x^{4} [/tex]-3x²-4x) \ (x²-1)²
= xg(x) \ (x²-1)²
3) x+2 + (x+2) \ x²-1
= (x³-x+2x²-2+x+2) \ (x²-1)
= (x³+2x²) \ (x²-1)
= f(x)
voilà , j'espère que tu as compris !
