2ã解ï¼f(x)=x³+ax²+bxï¼f '(x)=3x²+2ax+bï¼å·²ç¥ï¼
f '(1)=3+2a+b=0.........(1)ï¼f(1)=1+a+b=-12..........(2)
两å¼èç«æ±è§£å¾a=10ï¼b=-23.
3ãå·²ç¥arctan(y/x)=2lnâ(x²+y²)ï¼æ±dy/dx
解ï¼ä½å½æ°F(xï¼y)=arctan(y/x)-2lnâ(x²+y²)=0ï¼å
dy/dx=-(∂F/∂x)/(∂F/∂y)=-[-(y/x²)/(1+y²/x²)-2x/(x²+y²)]/[(1/x)/(1+y²/x²)-2y/(x²+y²)]=(y+2x)/(x-2y)
4ãæ±æéxâ0lim{1/x-1/[(e^x)-1]}
解ï¼åå¼=xâ0lim[(e^x)-1-x]/{x[(e^x)-1]}=xâ0lim[(e^x-1)]/[(e^x)-1+xe^x]
=xâ0lim[(e^x)/(2e^x+xe^x)=1/2
5ãå·²ç¥F(xï¼y)=y⁵+2y-x-3x⁷=0ï¼æ±dy/dx(x=0ï¼y=0)
解ï¼dy/dx=-(∂F/∂x)/(∂F/∂y)=(1+21x⁶)/(5y⁴+2)â£ãx=0ï¼y=0ã=1/2
6ãå·²ç¥y=xsin2xï¼æ±y'''
解ï¼y'=sin2x+2xcos2xï¼y''=2cos2x+2cos2x-4xsin2x=4cos2x-4xsin2xï¼
y'''=-8sin2x-4sin2x-8xcos2x=-12sin2x-8xcos2x.
7ãå·²ç¥xâ0lim[(ax+2sinx)/x]=2ï¼æ±a
解ï¼xâ0lim[(ax+2sinx)/x]=xâ0lim(a+2cosx)=2ï¼æ
a=0ãåé¢xââå¯è½æéã
8ãæ±æéxâÏ/2lim[(lnsinx)/(Ï-2x)²
解ï¼åå¼=xâÏ/2lim(cotx)/[-4(Ï-2x)]=xâÏ/2lim[-(csc²x)/8]=-1/8
9ãå·²ç¥y=∛[(x+1)(2x+1)/(x+2)(5-3x)]ï¼æ±y'
解ï¼ä¸¤è¾¹å对æ°å¾ln=(1/3)[ln(x+1)+ln(2x+1)-ln(x+2)-ln(5-3x)]
两边对xæ±å¯¼å¾y'/y=(1/3)[1/(x+1)+2/(2x+1)-1/(x+2)+3/(5-3x)]
æ
y'=(1/3)[1/(x+1)+2/(2x+1)-1/(x+2)+3/(5-3x)]y
=(1/3)[1/(x+1)+2/(2x+1)-1/(x+2)+3/(5-3x)]{∛[(x+1)(2x+1)/(x+2)(5-3x)]}
10ãy=xarccosx-â(1-x²)ï¼æ±dy
解ï¼dy=[arccosx-1/â(1-x²)+x/â(1-x²)]dx=[arccosx+(x-1)/â(1-x²)]dx
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