数学微积分题目：如果y=(1/u)(du/dx)，问dy/dx等于什么？ u=f（x,y,z）,求du/dx——du/dx是什么意思？...
\u6c42\u5fae\u5206\u65b9\u7a0bdy/dx=1/(x+y)\u7684\u901a\u89e3dy/dx=1/(x+y)
dx/dy=x+y
x'-x=y
x=e^-\u222b-dy\u00b7[\u222be^(\u222b-dy)\u00b7ydy+C]
=e^y\u00b7[\u222b(e^-y)\u00b7ydy+C]
=e^y\u00b7[-\u222byd(e^-y)+C]
=e^y\u00b7[-y\u00b7e^-y+\u222be^-ydy+C]
=e^y\u00b7[(-y-1)e^-y+C]
=Ce^y-y-1
\u6269\u5c55\u8d44\u6599\uff1a
\u5fae\u5206\u65b9\u7a0b\u7814\u7a76\u7684\u6765\u6e90\uff1a\u5b83\u7684\u7814\u7a76\u6765\u6e90\u6781\u5e7f\uff0c\u5386\u53f2\u4e45\u8fdc\u3002\u725b\u987f\u548cG.W.\u83b1\u5e03\u5c3c\u8328\u521b\u9020\u5fae\u5206\u548c\u79ef\u5206\u8fd0\u7b97\u65f6\uff0c\u6307\u51fa\u4e86\u5b83\u4eec\u7684\u4e92\u9006\u6027\uff0c\u4e8b\u5b9e\u4e0a\u8fd9\u662f\u89e3\u51b3\u4e86\u6700\u7b80\u5355\u7684\u5fae\u5206\u65b9\u7a0by'=f(x)\u7684\u6c42\u89e3\u95ee\u9898\u3002
\u5f53\u4eba\u4eec\u7528\u5fae\u79ef\u5206\u5b66\u53bb\u7814\u7a76\u51e0\u4f55\u5b66\u3001\u529b\u5b66\u3001\u7269\u7406\u5b66\u6240\u63d0\u51fa\u7684\u95ee\u9898\u65f6\uff0c\u5fae\u5206\u65b9\u7a0b\u5c31\u5927\u91cf\u5730\u6d8c\u73b0\u51fa\u6765\u3002\u725b\u987f\u672c\u4eba\u5df2\u7ecf\u89e3\u51b3\u4e86\u4e8c\u4f53\u95ee\u9898\uff1a
\u5728\u592a\u9633\u5f15\u529b\u4f5c\u7528\u4e0b\uff0c\u4e00\u4e2a\u5355\u4e00\u7684\u884c\u661f\u7684\u8fd0\u52a8\u3002\u4ed6\u628a\u4e24\u4e2a\u7269\u4f53\u90fd\u7406\u60f3\u5316\u4e3a\u8d28\u70b9\uff0c\u5f97\u52303\u4e2a\u672a\u77e5\u51fd\u6570\u76843\u4e2a\u4e8c\u9636\u65b9\u7a0b\u7ec4\uff0c\u7ecf\u7b80\u5355\u8ba1\u7b97\u8bc1\u660e\uff0c\u53ef\u5316\u4e3a\u5e73\u9762\u95ee\u9898\uff0c\u5373\u4e24\u4e2a\u672a\u77e5\u51fd\u6570\u7684\u4e24\u4e2a\u4e8c\u9636\u5fae\u5206\u65b9\u7a0b\u7ec4\u3002\u7528\u53eb\u505a\u201c\u9996\u6b21\u79ef\u5206\u201d\u7684\u529e\u6cd5\uff0c\u5b8c\u5168\u89e3\u51b3\u4e86\u5b83\u7684\u6c42\u89e3\u95ee\u9898\u3002
∂z/∂x\uff1a\u662f\u504f\u5bfc = partial differentiation\uff1b
dz/dx\uff1a\u662f\u5168\u5bfc = total differentiation\u3002
\u5bf9\u4e8e\u5168\u5bfc\uff0c\u624d\u6709\u5168\u5fae\u5206\uff1a
dz = (∂z/∂x)dx + (∂z/∂y)dy\u3002
∂u/∂x=f1'*[∂(x/y)/∂x]+f2'*[∂(y/z)/∂x]=f1'/y+f2'*0=f1'/y\uff1b
∂u/∂y=f1'*[∂(x/y)/∂y]+f2'*[∂(y/z)/∂y]=-(x/y²)f1'+(f2'/z)\uff1b
∂u/∂z=f1'*[∂(x/y)/∂z]+f2'*[∂(y/z)/∂z]=f1'*0-(y/z²)f2'=-(y/z²)f2'\uff1b
\u6269\u5c55\u8d44\u6599\uff1a
\u4e00\u4e00\u578b\u9501\u94fe\u6cd5\u5219
\u5728\u4e2d\u95f4\u53d8\u91cf\u53ea\u6709\u4e00\u4e2a\u65f6\uff0c\u5982z=f(u,x)\uff0c\u5b83\u5728\u76f8\u5e94\u70b9\u6709\u8fde\u7eed\u5bfc\u6570\uff0c\u5219\u53ef\u5f97\u4e00\u4e00\u578b\u5168\u5bfc\u6570\u9501\u94fe\u6cd5\u5219\uff0c\u5373\uff1a [1]
\u4e8c\u4e00\u578b\u9501\u94fe\u6cd5\u5219
\u8bbeu=u(x)\u3001v=v(x)\u5728x\u53ef\u5bfc\uff0cz=f(u,v)\u5728\u76f8\u5e94\u70b9(u,v)\u6709\u8fde\u7eed\u504f\u5bfc\u6570\uff0c\u5219\u590d\u5408\u51fd\u6570z=f(u(x),v(x))\u5728x\u53ef\u5bfc\uff0c\u4e14\u6709\uff1a
\u8bc1\u660e\uff1a\u5bf9\u4e8e\u81ea\u53d8\u91cfx\u7684\u8be5\u53d8\u91cf\u25b3x\uff0c\u53d8\u91cfu=u(x)\u3001v=v(x)\u7684\u6539\u53d8\u91cf\u25b3u,\u25b3v\uff0c\u8fdb\u4e00\u6b65\u6709\u51fd\u6570\u7684\u8be5\u53d8\u91cf\u25b3z\uff0c\u56e0\u4e3a\u51fd\u6570z=f(u,v)\u53ef\u5fae\uff0c\u5373\u6709
\u5bf9\u4e0a\u5f0f\u5de6\u53f3\u4e24\u7aef\u540c\u9664\u25b3x\uff0c\u5f97\u5230\uff1a
\u53c8\u56e0\u4e3au=u(x)\u3001v=v(x)\u53ef\u5bfc\uff0c\u5f53
\u65f6\uff0c\u5bf9\u4e0a\u5f0f\u5de6\u53f3\u4e24\u7aef\u540c\u65f6\u53d6\u6781\u9650\uff0c\u5219\u6709\uff1a
\u8bc1\u660e\u5b8c\u6bd5\u3002
\u53c2\u8003\u8d44\u6599\uff1a\u767e\u5ea6\u767e\u79d1-\u5168\u5bfc\u6570
求解如下：
绛旓細澶т竴鏁板姹寰Н鍒,寰堝ソ閲囩撼鐨 姹y=1+x涔樹互e鐨剏娆℃柟鐨勫井鍒 鎴戞潵绛 1涓洖绛 #鐑# 鐢熸椿涓湁鍝簺鎴愮樉椋熺墿?閲嶈繑2006REWT 2016-11-13 路 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 閲嶈繑2006REWT 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 閲囩撼鏁:802 鑾疯禐鏁:10410 姣曚笟浜庡箍瑗垮ぇ瀛︾幆澧冨伐绋嬩笓涓,纭曞＋瀛︿綅,瀵瑰彛涓撲笟宸ヤ綔3骞 鍚慣A鎻愰棶 绉佷俊TA ...
绛旓細姹俷闃跺鏁版病鏈夌粺涓鐨勫叕寮忥紝閮介渶瑕佺粨鍚棰樼洰鏉ュ叿浣撳垎鏋愮殑 锛1锛墆=e^(ax+b)y'=a*e^(ax+b)y''=a^2*e^(ax+b)...y^(n)=a^n*e^(ax+b)锛2锛y=(1-x)/(1+x)=(2-1-x)/(1+x)=2/(1+x)-1=2*[(1+x)^(-1)]-1 y'=2*(-1)*(1+x)^(-2)y''=2*(-1)*(-2...
绛旓細y=[(1/x)(a^x-1)/(a-1)]^(1/x)鍒欙細lny=(1/x)[ln(a^x-1)-ln(a-1)-lnx]lim[x鈫+鈭瀅 lny =lim[x鈫+鈭瀅 [ln(a^x-1)-ln(a-1)-lnx]/x 娲涘繀杈炬硶鍒 =lim[x鈫+鈭瀅 [a^xlna/(a^x-1)-1/x]=lim[x鈫+鈭瀅 [(a^x-1+1)lna/(a^x-1)-1/x]=lim[x鈫+...
绛旓細Y=lnX 鏈潵姹傚浠ュ悗灏辨槸1/X 鑰宔鐨2x姝ゆ柟 涓闃跺鏁颁负2涔樹互e鐨2x娆 鎵浠ヤ簩闃跺鏁颁负4涔樹互e2x娆℃柟
绛旓細鍥炵瓟锛氬垎鍒 x瀵箃姹傚绛変簬sin t y瀵箃姹傚绛変簬cost dy/dx=(y瀵箃鐨勫鏁)/(x瀵箃鐨勫鏁) 鏄笉鏄笉瀹绉垎鐨勫彉涓婁笅闄愭眰瀵间笉浼氭眰?
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绛旓細浠庡箍涔変笂璇达紝鏁板鍒嗘瀽鍖呮嫭寰Н鍒銆佸嚱鏁拌绛夎澶氬垎鏀绉戯紝浣嗘槸鐜板湪涓鑸凡涔犳儻浜庢妸鏁板鍒嗘瀽鍜屽井绉垎绛夊悓璧锋潵锛屾暟瀛﹀垎鏋愭垚浜嗗井绉垎鐨勫悓涔夎瘝锛涓鎻愭暟瀛﹀垎鏋愬氨鐭ラ亾鏄寚寰Н鍒哰2]銆備竴鍏冨井鍒 鎶樺彔瀹氫箟 璁惧嚱鏁板湪鏌愬尯闂村唴鏈夊畾涔夛紝鍙+ 螖x鍦ㄦ鍖洪棿鍐呫濡傛灉鍑芥暟鐨勫閲徫y = f(+ 螖x) 鈥 f(锛夊彲琛ㄧず涓 螖...
绛旓細绾挎т唬鏁板湪鏌愮鎰忎箟涓婃槸鈥滃畬鍏ㄨВ鍐充簡鈥,浣濡傛灉浣犺浣犵殑鏂圭▼涓湁涓嶅悓鎸囨暟,閭ｅ畠浠氨鏄椤瑰紡銆傝繖灏辨槸浠ｆ暟鍑犱綍:鍦ㄥ椤瑰紡闆堕泦鐨勫嚑浣曞拰澶勭悊杩欎簺鏂圭▼鐨勪唬鏁拌繍绠椾箣闂磋繘琛岃浆鎹 鍦ㄦ湰鏂囦腑,涓涓厜婊戜唬鏁扮皣(绠鍖栦负鈥滅皣鈥)鏄竴涓嚑浣曠┖闂碭,鐢卞椤瑰紡鐨勯浂闆嗙粰鍑,寰楀埌鐨勭┖闂存槸"鍏夋粦鐨",灏卞儚浣犱滑鍦寰Н鍒涓鍒扮殑閭ｆ牱銆 浜岀淮...
绛旓細y=tanx/(1+cosx)y' = [(secx)^2(1+cosx)-tanx(-sinx)]/(1+cosx)^2 = [1+cosx+cosx(sinx)^2]/[(cosx)^2(1+cosx)^2],1,
绛旓細1 lim(x鈫3锛y鈫0)sin(xy)/tan(xy)=lim(x鈫3,y鈫0)sin(xy)'/tan(xy)'=lim(x鈫3锛寉鈫0)xcos(xy)/[x/(cos(xy))^2]=lim(x鈫3锛寉鈫0)cos(xy)=1 2 lim(x鈫0,y鈫2) (1+sin(xy))^(y/x)=lim(x鈫0,y鈫2)锛1+sin(xy))^(y/sin(xy))^(sin(xy)/x)=lim(x...
