求函数y=e^-x.cos.3x的导数怎么求啊？？ y=e的-xcos3x的导数是多少?
\u6c42y=e^-x cos3x\u7684\u5bfc\u65702\u4e58\u4ee5e^2x\u4e58\u4ee5cos3x\u51cf\u53bbe^2x\u4e58\u4ee53\u4e58\u4ee5Sin3x\uff08y=e^x\u7684\u5bfc\u51fd\u6570\u662f\u5b83\u672c\u8eaby=e^x\uff0c\u800cy=e^2x\u7684\u5bfc\u51fd\u6570\u5c31\u662fy=e^2x\u8fd8\u8981\u4e58\u4ee5\u5185\u51fd\u65702X\u7684\u5bfc\u65702\uff0c\u5373\u5f97\u5230y=e^2x\u7684\u5bfc\u51fd\u6570\u662f2e^2x\uff09
\u539f\u51fd\u6570\u7684\u5bfc\u6570=e^x+cos3x-3xsin3x
\u628a(0,1)\u7684\u6a2a\u5750\u6807\u5e26\u5165\u5bfc\u6570\u5f97
\u5207\u7ebf\u7684\u659c\u7387k=e^0+cos0+0=1+1=2
\u6240\u4ee5\u5207\u7ebf\u65b9\u7a0b\u4e3ay-1=2(x-0)
\u5316\u7b80\u5f97y=2x+1
y=e^(-x)*cos3x
y'=-e^(-x)cos3x+e^(-x)*(-3sin3x)=-e^(-x)（cos3x+3sin3x）
(e^x的导数是e^x，cosx的导数为-sinx）
然后你可以根据情况考虑是否化简
中间两个点什么？如果题目是y=e^-x.cos3x的话那求导 y'=(cos3x一sin3x)/e^x 分子可用辅助角公式化同名，会吧？亲。
绛旓細棣栧厛璇疯浣忓叕寮廩f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)y=e^(-x)*cos3x y'=-e^(-x)cos3x+e^(-x)*(-3sin3x)=-e^(-x)锛坈os3x+3sin3x锛(e^x鐨勫鏁版槸e^x锛cosx鐨勫鏁颁负-sinx锛夌劧鍚庝綘鍙互鏍规嵁鎯呭喌鑰冭檻鏄惁鍖栫畝
绛旓細=2[e^(-x)]'sinx+2e^(-x)(sinx)'=-2e^(-x)sinx+2e^(-x)cosx =2e^(-x)(cosx-sinx)涓句緥 渚嬪锛y=x^3+3x^2+7x+9鐨勫鏁颁负y=3x^2+6x+7锛屼簩闃跺鏁板嵆y=3x^2+6x+7鐨勫鏁颁负y=6x+6锛屼笁闃跺鏁板嵆y=6x+6鐨勫鏁颁负y=6銆傜敱姝ゅ彲鎺ㄥ箍鍒皀闃跺鏁帮紝鍗冲皢鍘鍑芥暟杩涜n娆℃眰瀵笺備笁娆″嚱鏁...
绛旓細dy = -e^(-x)(cos3x-3sin3x)dx 姹俤y 灏辩浉褰撲簬姹傚涔嬪悗 涔樹笂涓涓猟x
绛旓細y'=-e^(-x)*(x+1)^(1/3)+e^(-x)/[3(x+1)^(2/3)] x=0鏃讹紝k1=y'|x=0=-1+1/3=-2/3, 鍒囩嚎涓y=(-2/3)x+1,娉曠嚎涓簓=1.5x+1; x=-1鏃讹紝k2=y'|x=-1=涓嶅瓨鍦紝 鍒囩嚎涓簒=-1,娉曠嚎涓簓=0
绛旓細y=e^(-x)路cos(3-x)y'=e^(-x)路(-x)'路cos(3-x)-e^(-x)路sin(3-x)路(3-x)'=-e^(-x)路cos(3-x)+e^(-x)路sin(3-x)鈭磀y=[e^(-x)路sin(3-x)-e^(-x)路cos(3-x)]dx =鈭2e^(-x)sin(3-x-¼蟺)dx ...
绛旓細2涔樹互e^2x涔樹互cos3x鍑忓幓e^2x涔樹互3涔樹互Sin3x锛y=e^x鐨勫鍑芥暟鏄畠鏈韩y=e^x锛岃寉=e^2x鐨勫鍑芥暟灏辨槸y=e^2x杩樿涔樹互鍐呭嚱鏁2X鐨勫鏁2锛屽嵆寰楀埌y=e^2x鐨勫鍑芥暟鏄2e^2x锛
绛旓細杩欎袱涓兘鏄井鍒嗗熀鏈殑娉曞垯锛屽仛杩欓亾棰樻椂閮戒細鐢ㄥ埌銆y=e^(-x)cos(3-x)dy/dx = {d[e^(-x)]/dx}cos(3-x)+ e^(-x){d[cos(3-x)]/dx}锛堜箻绉硶鍒欙級= -e^(-x)cos(3-x)+ e^(-x)(-1)(-sin(3-x))锛堥摼寮忔硶鍒欙級= e^(-x)[-cos(3-x)+sin(3-x)]
绛旓細鈭礵y/dx=y'=[e^(-x)]'cos(3-x)+e^(-x)[cos(3-x)]'=-e^(-x)cos(3-x)+e^(-x)(sin(3-x)=[-cos(3-x)+(sin(3-x)]e^(-x)鈭磀y=e^(-x)[-cos(3-x)+(sin(3-x)]dx
绛旓細杩欎袱涓兘鏄井鍒嗗熀鏈殑娉曞垯锛屽仛杩欓亾棰樻椂閮戒細鐢ㄥ埌銆y=e^(-x)cos(3-x)dy/dx = {d[e^(-x)]/dx}cos(3-x) + e^(-x){d[cos(3-x)]/dx}锛堜箻绉硶鍒欙級= -e^(-x)cos(3-x) + e^(-x)(-1)(-sin(3-x))锛堥摼寮忔硶鍒欙級= e^(-x)[-cos(3-x)+sin(3-x)]
绛旓細寰垎dy=[-e鐨勶紙-x锛cos(3-x)+e鐨勶紙-x锛塻in锛3-x锛塢dx 瑕佹兂姹傚鐨勮瘽灏辩洿鎺ユ妸dx绉诲埌鍓嶈竟鍘诲氨濂戒簡.,3,-1/2鍊嶇殑e鐨勶紙-x锛夋鏂逛箻浠ャ攃os(3-x)-sin(3-x)銆,0,
