的三积号达 高等数学
高 等 数 学
和、差、积、商的求导法 定理 如果函数(x),(x)在点x处可导, 则它们的和、差、积、商(分母不为零)在点x处也可导 (1)[(x)±w(x)]=t(x)±v(x); (2)[v(x)v(x)=u(x)v(x)+(x)v(x); (3)y=21(x)v(x)-u(x)v(x) (v(x)≠0) 1(x 1(x
一、和、差、积、商的求导法 定理 如果函数u(x), v(x)在点x处可导, (1)[u(x) v(x)] = u (x) v (x); 则它们的和、差、积、商(分母不为零)在点x处也可导, (2)[u(x)v(x)] = u (x)v(x) +u(x)v (x); ( ( ) 0). ( ) ( ) ( ) ( ) ( ) ] ( ) ( ) (3)[ 2 − = v x v x u x v x u x v x v x u x
证明(()[u(x)±v(x分=(x)(); 条件:f(x)=l(x)+v(x)结论:f(x)=(x)+p(x) 而函数(x)v(x)在点x处可导(x)+/(x) 由导数定义有: f(x)=lim f(x+h-f()=lim (x+h)-(x),v(x+h)-vx) h->0 h-0 h [(x+h)+v(x+h)-[v(x)+y(x) Im h->0 h 这表示,函数f(x)在点X处也可导,且f(x)=(x)+(x) 以上结果简单地写成:(+y)=+v 类似地可得证明(2):(l-y)=l-y
证明（1）： f (x) = u(x) + v(x) 由导数定义有： 这表示，函数 f (x) 在点x 处也可导，且 ( ) ( ) ( ) ' ' ' f x = u x + v x 以上结果简单地写成： ' ' ' (u −v) = u −v ' ' ' (u + v) = u + v 类似地可得证明（2）： 条件： 而函数u(x), v(x)在点x处可导 ( ) ( ) ( ) ' ' ' 结论： f x = u x + v x h f x h f x f x h ( ) ( ) ( ) lim 0 ' + − = → h u x h v x h u x v x h [ ( ) ( )] [ ( ) ( )] lim 0 + + + − + = → ( ) ( ) ' ' = u x + v x 0 ] ( ) ( ) ( ) ( ) lim[ h v x h v x h u x h u x h + − + + − → = (1)[u(x) v(x)] = u (x) v (x);
(3)/(r) u(xvey-u(xv(x (v(x)≠0) v(x 1(x 证(3)设f(x) ulr ,(v(x)≠0), f(r)=lim/(x+h)-f(r) h→0 u(x+h u(r) vr+ h) v(x) h→>0 lim u(x+ h)v(x)=u(x)v(x+h) h→>0 (x+ hv(x)h lim u(x+h)v(x)+u(x)v(x)=u(xv(x)u(x)v(x+h) h→>0 v(x+h)v(xh
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 − + + = → v x h v x h u x h v x u x v x u x v x u x v x h h ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lim 0 + + + − − + = → ( ( ) 0). ( ) ( ) ( ) ( ) ( ) ] ( ) ( ) (3)[ 2 − = v x v x u x v x u x v x v x u x
lim lu(x+h)-u(x)lv(x)-u(e)v(x+h)-v(x) h→>0 v(x+hv(x )h u(x+h)-u(x) v(x)-u(r) v(x+h-v(x) lim- h h→0 v(x+hv(x) u(xv(x)-u(x)v(r) lv(x)l ∫(x)在x处可导
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