第二节函数的和、差、积 商的求导法则 四一、和、差、积、商的求导法则 巴二、例题分析 小结思考题
、和、差、积、商的求导法则 中定理如果函数x,(x在点处可导则它 们的和、差、积、商分母不为零在点x处也 可导,并且 (1)[u(x)±v(x)=u'(x)±v(x) 工工工 (2)u(x)v(x)=u'(x)v(x)+u(x)y(x); ux ), u(xv(x-u(x)v(x) (3)= (v(x)≠0) v() v2(x) 上页
一、和、差、积、商的求导法则 定理 可 导 并 且 们的和、差、积、商 分母不为零 在 点 处 也 如果函数 在 点 处可导 则 它 , ( ) ( ), ( ) , x u x v x x ( ( ) 0). ( ) ( ) ( ) ( ) ( ) ] ( ) ( ) (3)[ (2)[ ( ) ( )] ( ) ( ) ( ) ( ); (1)[ ( ) ( )] ( ) ( ); 2 − = = + = v x v x u x v x u x v x v x u x u x v x u x v x u x v x u x v x u x v x
证(1)、(2)略 证(3)设∫(x=(x)(v(x)≠0, vlx f(x)=lim /(+ h)-f(r) h→>0 h (x+h)(x) = lim v(x+h) v(x) h- 0 h li u(x+hv(r)-u(r)v(x+h) h→>0 v(x+hv(x)h 上页
证(3) , ( ( ) 0), ( ) ( ) ( ) = v x v x u x 设 f x h f x h f x f x h ( ) ( ) ( ) lim 0 + − = → v x h v x h u x h v x u x v x h h ( ) ( ) ( ) ( ) ( ) ( ) lim 0 + + − + = → h v x u x v x h u x h h ( ) ( ) ( ) ( ) lim 0 − + + = → 证(1)、(2)略
u(thu(xlv(x-u(xlv(x+h)-v(x h→>0 v(+hv(r)h u(x+h)-u(x) v()-u(r) v(x+h-v(x) =lim h h→0 v(+ hv(x) u(x)v(x)-u(rv(x) v(x)l f(x)在x处可导 上页
v x h v x h u x h u x v x u x v x h v x h ( ) ( ) [ ( ) ( )] ( ) ( )[ ( ) ( )] lim 0 + + − − + − = → ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lim 0 v x h v x h v x h v x v x u x h u x h u x h + + − − + − = → 2 [ ( )] ( ) ( ) ( ) ( ) v x u x v x − u x v x = f (x)在x处可导
推论 中()②(x=∑fx; i=1 i=1 (2) Cf(r=c(r) 庄(3)m(o)=f(x)(x)-f(x) i=1 +…+f1(x)f2(x)…fm(x) =∑Ⅱf(x)/(x); i=1k=1 ≠i 上页
推论 (1) [ ( )] ( ); 1 1 = = = n i i n i fi x f x (2) [Cf (x)] = Cf (x); ( ) ( ); ( ) ( ) ( ) (3) [ ( )] ( ) ( ) ( ) 1 1 1 2 1 2 1 = + + = = = = n i n k i k i k n n n i i f x f x f x f x f x f x f x f x f x