Question

求数学大神第二问

Answer 1

当x≠0时，f'(x)={[g'(x)+sinx]*x-[g(x)-cosx]}/x^2
当x=0时，f'(0)=lim(t->0) [f(t)-f(0)]/t
=lim(t->0) {[g(t)-cost]/t-g'(0)}/t
=lim(t->0) [g(t)-cost-g'(0)t]/t^2
=lim(t->0) [g'(t)+sint-g'(0)]/(2t)
=lim(t->0) [g''(t)+cost]/2
=[g''(0)+1]/2
因为lim(x->0) f'(x)=lim(x->0) {[g'(x)+sinx]*x-[g(x)-cosx]}/x^2
=lim(x->0) {[g''(x)+cosx]*x+g'(x)+sinx-g'(x)-sinx}/(2x)
=lim(x->0) [g''(x)+cosx]/2
=[g''(0)+1]/2
=f'(0)
即f(x)在x=0点处一阶连续可导
综上所述，f'(x)={[g'(x)+sinx]*x-[g(x)-cosx]}/x^2