Question

微积分计算题解答

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Answer 1

Answer 2

dy/dx=3cos2x*(2x)'+4e^x
dy/dx=6cos2x+4e^x
dy=(6cos2x+4e^x)dx

(2) lim(1-2/x)^x=lim[(1-1/(x/2))^(x/2)]^2
=lim[(1-1(x/2))^(-x/2)]^(-2)
=e^(-2)

(3)f'(x)=6x^2-6=0 x=1 or x=-1 f''(x)=12x
f''(1)=12>0 x=1取极小值，为f(1)=2*1^3-6*1+1=2-6+1=-3
f''(-1)=-12<0 取极大值，为f(-1)=-2+6+1=5

(4)(1+2x^2)/(1+x^2)=(2(x^2+1)-1)/(1+x^2)=2-1/(1+x^2)
原式=2x-arctanx+C 附：arctanx'=1/(1+x^2)

Answer 3

①dy/dx=3cos(2x) * (2x)' + 4e^x = 6cos2x + 4e^x
dy=(6cos2x + 4e^x)dx
②lim(x->∞)(1-2/x)^x
=lim(x->∞)[1+1/(-x/2)]^x
=lim(x->∞)[1+1/(-x/2)]^[(-x/2)*(-2)]
=e^(-2)
=1/e²
③f(x)=2x³-6x+1
极值点的切线肯定是水平的,
f'(x)=6x²-6=0
x=±1,f(1)=-3,f(-1)=9
极值点(1,-3) (-1,9)
④原式=∫ [2-1/(1+x²)]dx
=2x - arctan(x) + C