求曲线y=lnx在(e，1)处的切线方程和法线方程 求过程 曲线Y=lnx在点(1,2)处的切线方程和法线方程
\u6c42\u66f2\u7ebfy=lnx\u5728\u70b9\uff08e\uff0c1\uff09\u5904\u7684\u5207\u7ebf\u65b9\u7a0b\u548c\u6cd5\u7ebf\u65b9\u7a0b\u3002\u6c42\u8fc7\u7a0b
\u8fc7\u7a0b\u5982\u56fe
y'=1/x
x=1 y'=1 \u5219\u5207\u7ebf\u659c\u7387\u4e3a1 \u6cd5\u7ebf\u659c\u7387\u4e3a-1
\u5207\u7ebf\u4e3a y-2=x-1 \u5373 y=x+2
\u6cd5\u7ebf\u4e3a y-2=-(x-1) \u5373 y=-x+3
则切线的斜率k=f'(e)=1/e
故切线方程为y-1=1/e(x-1)
法线的斜率k=-e
故切线方程为y-1=-e(x-1)
Y
y=x/e
绛旓細瑙ｇ敱棰樼煡鐐锛坋,1锛鍦鏇茬嚎y=lnx涓 鐢眣=lnx 姹傚寰梱鈥=1/x 鏁呭綋x=4鏃,y鈥=1/e 鍗冲垏绾跨殑鏂滅巼k=1/e 鏁呭垏绾挎柟绋嬩负y-1=1/e(x-e)鍗充负y=x/e.
绛旓細瑙ｇ敱棰樼煡鐐锛坋锛1锛鍦鏇茬嚎y=lnx涓 鐢眣=lnx 姹傚寰梱鈥=1/x 鏁呭綋x=4鏃讹紝y鈥=1/e 鍗冲垏绾跨殑鏂滅巼k=1/e 鏁呭垏绾挎柟绋嬩负y-1=1/e(x-e)鍗充负y=x/e銆
绛旓細瑙ｆ眰瀵y'=1/x 鍒欏垏绾跨殑鏂滅巼k=f'(e)=1/e 鏁呭垏绾挎柟绋嬩负y-1=1/e(x-1)娉曠嚎鐨勬枩鐜噆=-e 鏁呭垏绾挎柟绋嬩负y-1=-e(x-1)
绛旓細銆愮瓟妗堛戯細鍒囩嚎鏂圭▼锛歺-ey=0锛涙硶绾挎柟绋嬶細ex+y-(e2+1)=0.
绛旓細瑙ｏ細鈭祔=lnx锛屸埓y鈥= 1 x 锛屸埓鏇茬嚎y=lnx鍦鐐筂锛坋锛1锛夊鍒囩嚎鐨勬枩鐜噆= 1 e 锛屾洸绾縴=lnx鍦ㄧ偣M锛坋锛1锛夊鍒囩嚎鐨勬柟绋嬩负锛歽-1= 1 e 锛坸-e锛锛屾暣鐞嗭紝寰梱= 1 e x锛庢晠绛旀涓猴細y= 1 e x锛
绛旓細y鐨勫鏁=1/x=鍒囩嚎鐨勬枩鐜=1/e 鎵浠ュ垏绾跨殑鏂圭▼涓 y-1=k(x-e)y=(1/e)x 娉曠嚎鐨勬枩鐜=-e 娉曠嚎鐨勬柟绋嬩负y-1=-e(x-e)y=-ex+e^2+1
绛旓細y = lnx, y' = 1/x, k = y'(e) = 1/e 鍒囩嚎鏂圭▼ y - 1 = (1/e)(x-e),浠 x = 0锛 寰 y = 1 - 1 = 0,鍒囩嚎涓 y 杞翠氦鐐规槸 鍘熺偣 O(0, 0)
绛旓細y'=1/x 褰搙=e鏃 鏂滅巼y'=1/e 鍒囩嚎鏂圭▼涓y=x/e+b 灏锛坋锛1锛甯﹀叆鏂圭▼ 瑙ｅ緱b=0 鎵浠ュ垏绾挎柟绋嬩负y=x/e
绛旓細y鐨勫鏁=1/x=鍒囩嚎鐨勬枩鐜=1/e 鎵浠ュ垏绾跨殑鏂圭▼涓 y-1=k(x-e)y=(1/e)x 娉曠嚎鐨勬枩鐜=-e 娉曠嚎鐨勬柟绋嬩负y-1=-e(x-e)y=-ex+e^2+1
绛旓細瑙ｇ敱棰樼煡鐐锛坋,1锛鍦鏇茬嚎y=lnx涓 鐢眣=lnx 姹傚寰梱鈥=1/x 鏁呭綋x=4鏃,y鈥=1/e 鍗冲垏绾跨殑鏂滅巼k=1/e 鏁呭垏绾挎柟绋嬩负y-1=1/e(x-e)鍗充负y=x/e 鏇茬嚎鐨勬硶绾挎柟绋嬫眰瑙ｆ柟娉 璁炬洸绾挎柟绋嬩负y=f(x)鍦ㄧ偣(a,f(a))鐨勫垏绾挎枩鐜囦负f'(a)鍥犳娉曠嚎鏂滅巼涓-1/f'(a)鐢辩偣鏂滃紡寰楁硶绾挎柟绋嬩负锛歽=-(...
