数学微积分的题目求解答，过程要详细，答案要正确，好快的加分！ 大一微积分数学求解答，过程详细？
\u7ecf\u6d4e\u6570\u5b66\u57fa\u7840\u5fae\u79ef\u5206\u9898\u76ee \u6c42\u89e3\u7b54 \u770b\u56fe\u4ee5\u4e0b\u51e0\u9053\u9898\u7528\u5230\u4e86\u5982\u4e0b\u65b9\u6cd5
(tanx)'=\uff08secx)^2
(e^x )'=e^x
\u79ef\u5206a\u7684\u5bfc*b=a*b -\u79ef\u5206a*b\u7684\u5bfc
\u6362\u5143\u6cd5
\u671b\u91c7\u7eb3
\u4f60\u5199\u7684\u7b49\u5f0f\u662f\u5bf9\u7684\uff01\u4f30\u8ba1\u662f\u4f60\u8ba1\u7b97\u7684\u65f6\u5019\u758f\u5ffd\u4e86
f'(x)\u7684\u7b49\u5f0f\u4e2d\u3002\u4ee3\u5165x=0\uff0c\u5373\u6c42f'(0)\uff0c\u663e\u7136\u9664\u4e86\u7b2c\u4e00\u9879\u4e4b\u5916\uff0c\u540e\u9762\u6240\u6709\u9879\u5747\u542b\u6709\u56e0\u5f0fx
\u6240\u4ee5\uff0c\u9664\u7b2c\u4e00\u9879\u4e4b\u5916\uff0c\u5176\u4f59\u5747\u4e3a0
\u6240\u4ee5\uff0cf'(0)=1\u00d7(0+1)\u00d7(0+2)\u00d7\u2026\u2026\u00d7(0+2015)=2015!
z=xy x+y=1
z=x(1-x)=-x^2+x=-(x-1/2)^2+1/4
z'x=-2x+1
x=1/2,z'x=0, z最大=1/4
2
∫D∫(x^2+y^2)dxdy x=ρcosθ,y=ρsinθ x^2+y^2=ρ^2 dxdy=(1/2)(dρ)(θdθ)
D x^2+y^2=1,ρ=1 x^2+y^2=4 ρ=2 0<=θ<=π/2
∫∫(x^2+y^2)dxdy=∫[1,2]ρ^2dρ∫[0,π/2] (1/2)θdθ=∫[1,2](π^2/4)ρ^2dρ=(7π^2/12)
3
(cosnπ/3)^2 =(1/2)[1+cos(2nπ/3)]
-1/2<=cos2nπ/3<=1 周期为3函数，不收敛
级数(cosnπ/3)^2不收敛
4
(x^2-3y^2)dx +3xdy=0
如果是(x^2-3y)dx+3xdy=0
3xdy-3ydx= -x^2dx
3dy/x-3ydx/x^2=-dx
d(3y/x)=-dx
通解3y/x=-x+C
5
xy=xf(z)+yg(z)
x(y-f(z))=yg(z)
(y-f(z))-xf'x=yg'x
y(x-g(z))=xf(z)
x-g(z)-yg'y=xf'y
(y-f(z))=xdf/dx)+ydg/dx
(y-f(z))dx/dz=xdf/dz+ydg/dz
(x-g(z))=ydg/dy+xdf/dy
(x-g(z))dy/dz=ydg/dz+xdf/dz
[y-f(z)] dx/dz=[x-g(z)]dy/dz
[x-g(z)] (dz/dx) =[y-f(z)] (dz/dy)
绛旓細1銆佷袱杈瑰悓鏃舵眰瀵煎緱锛歔e^(x+y)](1+y')-3+4yy'=0 瑙ｅ緱锛歽'=[3-e^(x+y)]/[e^(x+y)+4y]2銆佲埆[-2鈫1] f(x) dx =鈭玔-2鈫0] x² dx + 鈭玔0鈫1] 2 dx =(1/3)x³ |[-2鈫0] + 2 =8/3 + 2 =14/3 3銆乴im[x鈫掆垶] (1+2/x)^(2x)=lim[x...
绛旓細棣栧厛 x 鈭(0,1) f(tx)dt=鈭(0,x) f(tx)d(tx)锛屼护tx=u锛屽嵆鍘熸柟绋嬪彲浠ュ寲绠涓 f(x)=鈭(0,x) f(u)d(u) +(1-x)*e^2x 锛屽绛夊紡涓よ竟姹傚鍙互寰楀埌锛宖 '(x)=f(x) +(1-2x)*e^2x 灏卞緱鍒颁簡涓闃剁嚎鎬у井鍒嗘柟绋嬶細f '(x) - f(x) =(1-2x)*e^2x 鐢卞叕寮忓彲浠ュ緱鍒板叾閫氳В涓...
绛旓細d(xlnx)/dx =lnx+x(lnx)'=lnx+x*1/x =lnx+1 2.瀵逛笂闈㈢殑绛夊紡涓よ竟姹傜Н鍒 鈭玠(xlnx)dx=鈭(lnx+1)dx 鍒欏洜涓烘眰瀵间笌绉垎浜掍负閫嗚繍绠楋紝鎵浠 鈭玠(xlnx)dx=xlnx 鍒檟lnx=鈭(lnx+1)dx=鈭玪nxdx+鈭1dx =鈭玪nxdx+x+c 鎵浠:鈭玪nxdx=xlnx-x+c ...
绛旓細1銆佺瓟妗堟槸cos(y+2)銆愯В鏋愩戜竴闃跺亸瀵兼暟zx=sin(y+2)鎵浠ワ紝浜岄樁鍋忓鏁颁负 zxy=cos(y+2)2銆佺瓟妗堟槸2/5 銆愯В鏋愩戝墠涓涓嚱鏁版槸濂囧嚱鏁锛岀Н鍒涓0 鎵浠ワ紝鍘熷紡=鈭(-1~1)x^4路dx=2/5
绛旓細4銆5涓ら鐢ㄤ簩鍏冨嚱鏁版瀬闄愮殑娲涘繀杈炬硶鍒 lim[f(x,y)/g(x,y)]=lim{[f'x(x,y)dx+f'y(x,y)dy]/[g'x(x,y)dx+g'y(x,y)dy]} 鍏朵腑锛屾瀬闄愮偣涓 x->x0, y->y0锛屼笖dx=x-x0, dy=y-y0 绗4棰橈細lim{[鈭(xy+4)-2]/(xy)} x->1, y->0 f(x,y)=鈭(xy+4)-2, ...
绛旓細鎬ヤ粈涔堟ワ紵杩欎箞绠鍗鐨勯涔熺敤鏉ユワ紝杩樻湁浠涔堝彲浠ヤ笉鎬ョ殑锛1锛(x+1)³/3-sinx+C锛2锛塯.e. =鈭玿[(1+cos2x)/2]dx (鐢ㄥ嶈鍏紡闄嶉樁)= x²/4+ (1/2)*鈭玿cos2xdx = x²/4+ (1/4)*鈭玿d(sin2x)= x²/4+ (1/4)*(xsin2x-鈭玸in2xdx)= x²/4+...
绛旓細,鍙互鐢ㄦ崲鍏冩硶绉垎銆備互涓嬫槸瑙ｇ瓟锛氳3cosx-sinx=a(cosx+sinx)+b(-sinx+cosx)锛屾眰寰梐=1锛宐=2銆傚垯(3cosx-sinx)/(cosx+sinx)=1+2(-sinx+cosx)/(cosx+sinx)鎵浠ュ師绉垎=鈭1+2(-sinx+cosx)/(cosx+sinx)dx =鈭1dx+2鈭(-sinx+cosx)/(cosx+sinx)dx =x+2ln|cosx+sinx|+C ...
绛旓細瑙ｏ細銆愭瘡涓棰樼洰鐨勫閫夌瓟妗堬紝鑷笂鑰屼笅4涓溾棆鈥濓紝椤哄簭缂栧彿涓篈銆丅銆丆銆丏銆戠涓棰橈紝C銆傗埖鈭(-1)^(n-1)/n^(3/2)<鈭1/n^(3/2)锛屽悗鑰呮敹鏁涖傜浜岄锛孋銆傗埖x=-1鏃讹紝鏄氦閿欑骇鏁帮紝婊¤冻鏉ヨ幈甯冨凹鍏瑰垽鍒硶瀹氱悊鐨勬潯浠讹紝鏀舵暃銆傜涓夐锛孊銆傗埖鏄氦閿欑骇鏁帮紝婊¤冻鏉ヨ幈甯冨凹鍏瑰垽鍒硶瀹氱悊鐨勬潯浠讹紝鏀舵暃銆...
绛旓細=2蟺鈭玔1,3]ln(r^2+1)d(r^2+1)/2 =蟺[(r^2+1)ln(r^2+1)|[1,3]-鈭玔1,3](r^2+1)d(ln(r^2+1))]=蟺[10ln10-2ln2-鈭玔1,3]d(r^2+1)]=蟺[10ln10-2ln2-(r^2+1)|[1,3]]=蟺(10ln10-2ln2-8)璇存槑:鈭玔a,b]f(x)dx琛ㄧず绉垎鍖洪棿涓篬a,b]琚Н鍑芥暟涓...
绛旓細1锛塪A/dt=鈭玡^(-3t)cos2tdt =1/2*鈭玡^(-3t)d(sin2t)=1/2*[e^(-3t)sin2t+3*鈭玡^(-3t)sin2tdt]=1/2*e^(-3t)sin2t-3/4*鈭玡^(-3t)d(cos2t)=1/2*e^(-3t)sin2t-3/4*[e^(-3t)cos2t+3鈭玡^(-3t)cos2tdt]=1/2*e^(-3t)sin2t-3/4*e^(-3t)cos2t-9...
