y=sin2xcos3x这个函数如何求导
y=(sinx)/x\u8fd9\u4e2a\u51fd\u6570\u5982\u4f55\u6c42\u5bfc\u5939\u903c\u6cd5\uff0c\u4e0d\u662f\u4ec0\u4e48\u5939\u89e3\u6cd5\u3002
\u662f\u6c42x\u8d8b\u8fd1\u4e8e0\u65f6\uff0c\u51fd\u6570y=(sinx)/x\u7684\u6781\u9650\u7684\u4e00\u79cd\u65b9\u6cd5\uff0c
\u5f53x\u8d8b\u8fd1\u4e8e0\u65f6\uff0c\u51fd\u6570y=(sinx)/x\u7684\u6781\u9650\u4e3a\uff1a1\u3002
\u5229\u7528\u5355\u4f4d\u5706\uff0c\u53ca\u4e09\u89d2\u51fd\u6570\u7684\u5b9a\u4e49\uff0c
\u56fe\u5f62\u7ed3\u5408\uff0c\u5f53x\u4e3a0\u5230\u03c0/2\u7684\u89d2\u65f6\uff0c\u6709\uff1asinx<=x<=tanx\uff0c
\u6240\u4ee5 cosx<=(sinx)/x<=1\uff1b
\u5f53x\u8d8b\u8fd1\u4e8e0\u65f6\uff0ccosx\u8d8b\u8fd1\u4e8e1\uff0c
\u6240\u4ee5\u5f53x\u8d8b\u8fd1\u4e8e0\u65f6\uff0c(sinx)/x\u7684\u6781\u9650\u4e3a1\u3002
\u8bbey=x^x,\u5219ln y=xln x,\u4e24\u8fb9\u9690\u51fd\u6570\u6c42\u5bfc\u5f97y'/y=ln x+x/x=ln x+1, \u5c06y=x^x\u4ee3\u5165\uff0c\u5f97y'=x^x(ln x+1).
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绛旓細閭ｅ氨鍏堢粰浣犱釜瀹氱悊澶嶅悎鍑芥暟f(g(x))鐨勫鏁颁负f'(g(x))*g'(x)鏈夎繖涓畾鐞嗗氨鍙互寰楀埌锛氬湪绗竴涓嚱鏁颁腑f(x)=x^2 g(x)=sinx鎵浠ュ鏁颁负2sinx*cosx=2sinxcosx鍚岀悊,鍦ㄧ浜屼釜鍑芥暟閲岄潰,f(x)=x^3 g(x)=cosx鎵浠ュ鏁颁负3cos^2x*-sinx...
绛旓細瑙ｏ細y=sin2xcos3x 浠(x)=y f(0)=0 f'(x)=2cos2xcos3x-3sin3xsin2x 鈭磃'(0)=2 鈭磋繃x=0澶勭殑鍒囩嚎鏂圭▼涓猴細y=2x
绛旓細y=sin2xcos3x =(1/2)sin(2x+3x)+(1/2)sin(2x-3x)=(1/2)sin5x-(1/2)sinx y(20)=(1/2)sin(5x+20*蟺/2)*5^20-(1/2)sin(x+20*蟺/2)=(1/2)*5^20sin(5x+10蟺)-(1/2)sin(x+10蟺)=(1/2)*5^20sin5x-(1/2)sinx ...
绛旓細銆傘傘
绛旓細鎴戜滑瑕佺敤鎹㈠簳鍏紡鏉ヨ浆鎹 sin(2x)cos(3x) == cos(2) 杩欎釜绛夊紡銆傞鍏堬紝鎴戜滑闇瑕佹槑纭崲搴曞叕寮忔槸浠涔堬紝骞剁悊瑙ｅ浣曞簲鐢ㄥ畠銆傛崲搴曞叕寮忛氬父鐢ㄤ簬瀵规暟鍑芥暟锛岀敤鏉ュ皢涓涓簳鏁扮殑瀵规暟杞崲涓哄彟涓涓簳鏁扮殑瀵规暟銆傚叕寮忎负锛歭og_b(a) = log_c(a) / log_c(b)锛屽叾涓 c 鍙互鏄换浣曟鏁帮紝涓嶇瓑浜1銆備絾鍦ㄦ湰棰樹腑锛...
绛旓細銆愯В绛斻cos3xsin2x=3cos2x−sin2x2sinxcosx=3−tan2x2tanx锛屼护t=tanx锛屽垯t=0锛屼笖3−tan2x2tanx=3−t22t=−2 t2−3t锛屽綋t>0鏃讹紝t2−3t>0锛屽彲寰−2t2−3t<0锛涘綋t<0鏃讹紝t2−3t<0锛屽彲寰−2t2−3t...
绛旓細鍑芥暟y锛漵in2x锛cos3x鐨勫懆鏈熸槸2蟺銆傝繃绋嬪涓嬶細鍥犱负cos3x鐨勬渶灏忔鍛ㄦ湡涓2蟺/3 sin2x鐨勬渶灏忔鍛ㄦ湡涓合 瀹冧滑鐨勬渶灏忔鍛ㄦ湡鐨勬渶灏忓叕鍊嶆暟涓2蟺 鎵浠2蟺鏄嚱鏁皔=cos3x+sin2x鐨勪竴涓懆鏈
绛旓細y'=(sin2x)'(cos3x)锛(sin2x)(cos3x)'=2cos2xcos3x锛3sin2xsin3x =2(cos2xcos3x锛峴in2xsin3x )锛峴in2xsin3x =2cos(2x+3x)锛(cosx-cos5x)/2 =2.5cos5x锛0.5cosx
绛旓細鈭sin2xcos3xdx=(cosx)/2-(cos5x)/10+C銆傦紙C涓虹Н鍒嗗父鏁帮級瑙ｇ瓟杩囩▼濡備笅锛氣埆sin2xcos3xdx =鈭1/2(sin(2x+3x)+sin(2x-3x))dx锛堢Н鍖栧拰宸級=1/2鈭玸in5xdx-1/2鈭玸inxdx =1/10鈭玸in5xd5x+1/2鈭玠cosx =(cosx)/2-(cos5x)/10+C ...
绛旓細鍏朵腑鐢ㄥ埌鍜屽樊鍖栫Н鍏紡鍜屾寮﹀嚱鏁皀闃舵眰瀵煎叕寮,鎰氭槯涓嶆噦鐨勫湴鏂瑰彲浠ョ户缁棶鎴
