Si j'ai bien compris l'expression de f(x) on part donc de
f(x) = x + (2/x) soit f(x),=,x,+,2,x,f(x),=,x,+, \frac{2}{,x, }f(x),=,x,+,,x,2
Il faut donc caculer f(2) et f(2+h)
f(2),=,2,+,2,2,,=,2,+,1,=,3f(2), =, 2,+, \frac{2}{,2,},= ,2,+,1,=,3f(2),=,2,+,,2,2,=,2,+,1,=,3
f(2+h),=,2+h,+,2,2+h,,=,,(2+h)2,+,2,2+h,=,,h2+4h+6,2+hf(2+h), =, 2+h,+, \frac{2}{,2+h,},=,\frac{,(2+h)^2,+,2,}{2+h},=,\frac{,h^2+4h+6,}{2+h}f(2+h),=,2+h,+,,2+h,2,=,2+h,(2+h)2,+,2,,=,2+h,h2+4h+6,
Donc f(2+h),−,f(2),=,,h2+4h+6,2+h,−3,=,h2+4h+6−3(h+2)2+hf(2+h),-,f(2),=,\frac{,h^2+4h+6,}{2+h},-3, =,\frac{h^2 + 4h + 6-3(h+2)}{2+h}f(2+h),−,f(2),=,2+h,h2+4h+6,,−3,=,2+hh2+4h+6−3(h+2)
f(2+h),−,f(2),=,h2+h2+h,=,h(h+1)h+2f(2+h),-,f(2),=, \frac{h^2 + h}{2+h}, =,\frac{h(h+1)}{h+2}f(2+h),−,f(2),=,2+hh2+h,=,h+2h(h+1)
Donc f(2+h),−,f(2)h,=,,h+1,h+2\frac{f(2+h),-,f(2)}{h},=,\frac{,h+1,}{h+2}hf(2+h),−,f(2),=,h+2,h+1,
Pour trouver le nombre dérivé il faut donc chercher limh→0(,h+1,h+2)\lim_{h \rightarrow 0} (\frac{,h+1,}{h+2})limh→0(h+2,h+1,)
