已知y=ln(1+x^2),在x=1处的微分dy=
dy=(2x)/(1＋x²)则：dy|(x=1)=1
绛旓細鐪嬩綘杩欐嫭鍙锋墦鐨勶紝骞虫柟鍦ㄤ粈涔堝湴鏂圭粨鏋滃彲涓嶄竴鏍凤紝鎬讳箣鍏堟眰鍑簓', 鐒跺悗dy=y'dx.濡傛灉骞虫柟鍦(1-x)锛鍒檡'=-2(1-x)/(1-x)^2=2/(x-1). 濡傛灉骞虫柟鏁翠釜寮忓瓙锛屽垯y'=2ln(1-x)/(x-1).
绛旓細z'x = 2x/(1+x^2+y^2) = 2/6 = 1/3 z'y = 2y/(1+x^2+y^2) = 4/6 = 2/3
绛旓細瑙ｏ細鍥犱负y锛濃厾n锛1锛2x锛2锛夛紝鎵浠ユ湁 dy锛濓紙1锛2x锛2锛夛紘dx锛忥紙1锛2x锛2锛夛紳4xdx锛忥紙1锛2x锛2锛夈
绛旓細1銆佸鏁ln(1-x)鐨勬嘲鍕掑叕寮忔槸锛歭n(1+x)=x-x^2\2+x^3\3-x^4\4+...+(-1)^(n-1)x^n\n+O(x^(n+1))2銆佸湪鏁板涓紝娉板嫆鍏紡鏄竴涓敤鍑芥暟鍦ㄦ煇鐐圭殑淇℃伅鎻忚堪鍏堕檮杩戝彇鍊肩殑鍏紡銆傚鏋滃嚱鏁拌冻澶熷厜婊戠殑璇锛屽湪宸茬煡鍑芥暟鍦ㄦ煇涓鐐圭殑鍚勯樁瀵兼暟鍊肩殑鎯呭喌涔嬩笅锛屾嘲鍕掑叕寮忓彲浠ョ敤杩欎簺瀵兼暟鍊煎仛绯绘暟鏋勫缓...
绛旓細y=ln(x+鈭氾紙x^2+1锛)鐨勫鏁颁负锛1/鈭氾紙x^2+1锛夈傝В绛旇繃绋嬪涓嬶細瀵煎嚱鏁 濡傛灉鍑芥暟鐨勫鍑芥暟鍦ㄦ煇涓鍖洪棿鍐呮亽澶т簬闆讹紙鎴栨亽灏忎簬闆讹級锛岄偅涔堝嚱鏁板湪杩欎竴鍖洪棿鍐呭崟璋冮掑锛堟垨鍗曡皟閫掑噺锛夛紝杩欑鍖洪棿涔熺О涓哄嚱鏁扮殑鍗曡皟鍖洪棿锛屽鍑芥暟绛変簬闆剁殑鐐圭О涓哄嚱鏁扮殑椹荤偣锛屽湪杩欑被鐐逛笂鍑芥暟鍙兘浼氬彇寰楁瀬澶у兼垨鏋佸皬鍊硷紙鍗...
绛旓細x^2+1)=(x-1)^2/(x^2+1),鍙互鐪嬪嚭y鏈夊敮涓鐨勫鏁颁负0鐨勭偣锛屽嵆x=1.鎴戜滑鐭ラ亾y鐨勫畾涔夊煙涓哄叏浣撳疄鏁帮紝浠庤屽畠鐨勬瀬鍊肩偣蹇呯劧鏄鏁颁负0鐨勭偣锛屽嵆x=1杩涓鐐广傚彟涓鏂归潰鎴戜滑鐭ラ亾y'>0锛屼粠鑰寉鏄崟璋冨鍔犲嚱鏁帮紙鍥犱负涓鐐逛笉褰卞搷鍗曡皟鎬),浠庤屽湪鎯熶竴鍙兘鍙栨瀬鍊煎緱x=1澶勪篃涓嶈兘鍙栨瀬鍊硷紝閭ｅ氨閫塂浜嗐
绛旓細y=ln[x+(1+x^2)^(1/2)],dy/dx={1/[x+(1+x^2)^(1/2)]}*銆恱+(1+x^2)^(1/2)銆戔={1/[x+(1+x^2)^(1/2)]}*銆1+x/(1+x^2)^(1/2)銆=1/(1+x^2)^(1/2)鎵浠 dx/dy=(1+x^2)^(1/2)
绛旓細y=ln(1+x)涓哄崟璋冮掑鍑芥暟 x=0鏃讹紝y=ln1=0 x=1鏃讹紝y=ln2 鈭磝鈭(0,1)鏃讹紝y鈭(0,ln2)鑰屽張x鈭(0,未)锛屸埓未鐨勫彇鍊艰寖鍥翠负0<未鈮1
绛旓細y=ln(x+鈭氾紙x^2+1锛)鐨勫鏁颁负锛1/鈭氾紙x^2+1锛夈傝В绛旇繃绋嬪涓嬶細瀵兼暟璁＄畻鐨勬ц川锛氫笉鏄墍鏈夌殑鍑芥暟閮芥湁瀵兼暟锛涓涓嚱鏁颁篃涓嶄竴瀹氬湪鎵鏈夌殑鐐逛笂閮芥湁瀵兼暟銆傝嫢鏌愬嚱鏁板湪鏌愪竴鐐瑰鏁板瓨鍦紝鍒欑О鍏跺湪杩欎竴鐐瑰彲瀵硷紝鍚﹀垯绉颁负涓嶅彲瀵笺傜劧鑰岋紝鍙鐨勫嚱鏁颁竴瀹氳繛缁紱涓嶈繛缁殑鍑芥暟涓瀹氫笉鍙銆傚浜庡彲瀵肩殑鍑芥暟f(x)锛...
绛旓細dy=(dy/dx)dx = {1/[x+鈹(x²+1)]}[1+x/鈹(x²+1)]dx = 1/鈹(x²+1) dx 鈹忚〃绀烘牴鍙 y+xy' - e^x = y'cosy y' = [e^x-y]/[x-cosy]dy = [e^x-y]/[x-cosy]dx
