Réponse:
k(1+h) = 5/(1+h)²
k(1) = 5
t(h) = [k(1+h)-k(1)]/h
t(h) = [5/(1+h)² - 5]/h
t(h) = [5-5(1+h)²]/[h(1+h)²]
t(h) = [5-5(h²+2h+1)]/[h(1+h)²]
t(h) = (-5h²-10h)/[h(1+h)²]
t(h) = (-5h-10)/(1+h)²
2)
t(h) tend vers -10 quand h tend vers 0 donc k(x) est derivable en 1 et k'(1) = -10
