六:解:
f(x) = [(x-a)^2]* φ(x),
f'(x) = 2*(x-a)*φ(x)+ [(x-a)^2]* φ'(x),
f''(x) = 2*φ(x) + 2*(x-a)*φ(x) + 2*(x-a)*φ'(x) + [(x-a)^2]* φ''(x),
so f''(a) = 2*φ(a)
七:
lim [F(1-cosx)]/tan(x^2) = lim {F'(1-cosx) *sinx/[sec(x^2)* (2x)]} = lim [F'(1-cosx)*cos(x^2) /2]
= F'(0)/2