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解答:
解
dy/dx=d(acosθ)/d(asin²θ)
=-asinθdθ/2asinθcosθdθ
=-1/2cosθ
dy/dx|(θ=π/4)=-1/2×(2/√2)=-√2/2
解
dy/dx=d(acosθ)/d(asin²θ)
=-asinθdθ/2asinθcosθdθ
=-1/2cosθ
dy/dx|(θ=π/4)=-1/2×(2/√2)=-√2/2
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根号里面都是平方,而且求根号后也是大于等于0。如果只在实数范围内考虑则(0,0)至少是极大值点。
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(4)
lim(x->∞) (x-arctanx)/(x+arctanx)
=lim(x->∞) (1-arctanx/x)/(1+arctanx/x)
=(1-0)/(1+0)
=1
(6)
(2x+1)/(2x-1) =1+2/(2x-1)
let
1/y =2/(2x-1)
lim(x->∞)[ (2x+1)/(2x-1)]^(3x)
=lim(x->∞)[ 1+2/(2x-1)]^(3x)
=lim(y->∞)[ 1+1/y]^(3(2y+1)/2)
=lim(y->∞)[ 1+1/y]^(3y+3/2)
=lim(y->∞)[ 1+1/y]^(3y)
=e^3
(8)
tan(x-α) =(tanx-tanα)/(1+tanx.tanα)
(tanx-tanα) =tan(x-α).(1+tanx.tanα)
lim(x->α) (tanx-tanα)/(x-α)
=lim(x->α) tan(x-α).(1+tanx.tanα)/(x-α)
=lim(x->α) (x-α).(1+tanx.tanα)/(x-α)
=lim(x->α) (1+tanx.tanα)
=1+(tanα)^2
lim(x->∞) (x-arctanx)/(x+arctanx)
=lim(x->∞) (1-arctanx/x)/(1+arctanx/x)
=(1-0)/(1+0)
=1
(6)
(2x+1)/(2x-1) =1+2/(2x-1)
let
1/y =2/(2x-1)
lim(x->∞)[ (2x+1)/(2x-1)]^(3x)
=lim(x->∞)[ 1+2/(2x-1)]^(3x)
=lim(y->∞)[ 1+1/y]^(3(2y+1)/2)
=lim(y->∞)[ 1+1/y]^(3y+3/2)
=lim(y->∞)[ 1+1/y]^(3y)
=e^3
(8)
tan(x-α) =(tanx-tanα)/(1+tanx.tanα)
(tanx-tanα) =tan(x-α).(1+tanx.tanα)
lim(x->α) (tanx-tanα)/(x-α)
=lim(x->α) tan(x-α).(1+tanx.tanα)/(x-α)
=lim(x->α) (x-α).(1+tanx.tanα)/(x-α)
=lim(x->α) (1+tanx.tanα)
=1+(tanα)^2
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