3.1.13.1.43.1.63.1.73.1.23.1.33.1.5例6求下列分段函数的导函数α, ≥0f(α) =α,<0解函数f(）由函数=≥o)以及y=（o)在=o处拼接而成.当>0时，f(α)=3c2当<0时，f'(α）=2α，当=0时(Aa)3f(O +△a)-f(O)limlim=0AaArA2-0+A→0+(A)2f(O +△α) -f(o)limlim=0AaAaAa-0A-0所以3c2,>0f'(α) =0,α=02,a<o返回全屏关闭退出I=I11/41
3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 ~ 6 ¦e©ã¼ê¼ê f(x) = x 3 , x > 0 x 2 , x < 0 ) ¼ê f(x) d¼ê y = x 3 (x > 0) ±9 y = x 2 (x < 0) 3 x = 0 ? © ¤. x > 0 , f 0 (x) = 3x 2 , x < 0 , f 0 (x) = 2x, x = 0 , lim ∆x→0+ f(0 + ∆x) − f(0) ∆x = lim ∆x→0+ (∆x) 3 ∆x = 0 lim ∆x→0− f(0 + ∆x) − f(0) ∆x = lim ∆x→0− (∆x) 2 ∆x = 0 ¤± f 0 (x) = 3x 2 , x > 0 0, x = 0 2x, x < 0 11/41 kJ Ik J I £ ¶ '4 òÑ
3.1.73.1.13.1.23.1.33.1.43.1.53.1.6定理1设f(α)在αo可导，则f()在Co连续.换句话说.如果函数在一点Co不连续，则显然在Co处不可导证明由已知条件，极限f(α) - f(co)= f'(o)limT→TO-o存在.所以f(α) - f(co)lim (f(α) 一 f(aco)) = limTTOTTO-cof(α) - f(co)lim (α - co)lim三T-→-oT→TO= f'(co) · 0 = 0.由此得lim f(α) = f(co),T→TO即,f(α)在 o连续返回全屏关闭退出II12/41
3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 ½n 1 f(x) 3 x0 , K f(x) 3 x0 ëY. é{`, XJ¼ê3: x0 ØëY, Kw,3 x0 ?Ø. y² d®^, 4 lim x→x0 f(x) − f(x0) x − x0 = f 0 (x0) 3. ¤± lim x→x0 (f(x) − f(x0)) = lim x→x0 f(x) − f(x0) x − x0 (x − x0) = lim x→x0 f(x) − f(x0) x − x0 lim x→x0 (x − x0) = f 0 (x0) · 0 = 0. dd lim x→x0 f(x) = f(x0), =, f(x) 3 x0 ëY. 12/41 kJ Ik J I £ ¶ '4 òÑ
3.1.13.1.73.1.23.1.33.1.43.1.53.1.6例 7 函数 f()=lαl在 =0处连续,但是在=0不可导证明当=0时A-1limf*(0) =AaAa-→0+Ac-0limf' (0) =1AcAa-→0-所以f(α)=α在α=0处不可导.注意，从图3.5可以看出，函数在=0有一个尖点，即在=0不光滑，所以没有切线图3.5返回全屏关闭退出I413/41
3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 ~ 7 ¼ê f(x) = |x| 3 x = 0 ?ëY, ´3 x = 0 Ø. y² x = 0 f 0 + (0) = lim ∆x→0+ ∆x − 0 ∆x = 1, f 0 −(0) = lim ∆x→0− −∆x − 0 ∆x = −1. ¤± f(x) = |x| 3 x = 0 ?Ø. 5¿, lã3.5 ±wÑ, ¼ê3 x = 0 kk:, =3 x = 0 Ø1w, ¤±vk. ✲ ✻ . . . x y ã 3.5 13/41 kJ Ik J I £ ¶ '4 òÑ
3.1.13.1.53.1.63.1.73.1.23.1.33.1.4函数的四则运算3.1.2定理2（函数的和差积商的导数）设f(α）和g(α）可导，则f(a)±g(a),f(a)g(a) 及 (当 g(a) ≠0 时)皆可导,并有T1° (f(α)±g(α)) = f(α)±g(α);2°(f(α) · g(α))' = f'(α)g(α) + f(c)g'(α)(f(α))f'(α)g(α) - f(α)g'(α)3°g(c)g2(αc)证明关于1°，直接由导数的定义和极限的运算即可证得.关于2°，利用下列恒等式f(α + △)g(α + ) - f(α)g(α)= f(α + Aα)g(α + △α) - f(α)g(α + α) + f(α)g(α + △) - f(α)g(α)返回全屏关闭退出14/41
3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 3.1.2 ¼êoK$ ½n 2 (¼êÚ Èûê) f(x) Ú g(x) , K f(x) ± g(x), f(x)g(x) 9 f(x) g(x) £ g(x) 6= 0 ¤, ¿k 1 ◦ (f(x) ± g(x))0 = f 0 (x) ± g 0 (x); 2 ◦ (f(x) · g(x))0 = f 0 (x)g(x) + f(x)g 0 (x); 3 ◦ f(x) g(x) 0 = f 0 (x)g(x) − f(x)g 0 (x) g 2(x) . y² 'u 1 ◦ , dê½ÂÚ4$=y. 'u 2 ◦ , |^ eð ª f(x + ∆x)g(x + ∆x) − f(x)g(x) = f(x + ∆x)g(x + ∆x) − f(x)g(x + ∆x) + f(x)g(x + ∆x) − f(x)g(x) 14/41 kJ Ik J I £ ¶ '4 òÑ