【题目】已知f(x)=(ex−1)(e2x−2)(e3x−3)(e4x−4) f(x)=(e^x-1)(e^{2x}-2)(e^{3x}-3)(e^{4x}-4)\\ f(x)=(ex−1)(e2x−2)(e3x−3)(e4x−4)求f′(0).f'(0).f′(0).
【解答】令ex=te^x=tex=t，则eax=(ex)a=tae^{ax}=(e^x)^a=t^aeax=(ex)a=ta，于是
f(x)=(t−1)(t2−2)(t3−3)(t4−4)f(x)=(t-1)(t^2-2)(t^3-3)(t^4-4)f(x)=(t−1)(t2−2)(t3−3)(t4−4)=[(t−1)(t2−2)]⋅[(t3−3)(t4−4)]=[(t-1)(t^2-2)]\cdot[(t^3-3)(t^4-4)]=[(t−1)(t2−2)]⋅[(t3−3)(t4−4)]=(t3−2t−t2+2)(t7−4t3−3t4+12)
=(t^3-2t-t^2+2)(t^7-4t^3-3t^4+12)
=(t3−2t−t2+2)(t7−4t3−3t4+12)=t10−4t6−3t7+12t3=\color{Blue}t^{10}-4t^6-3t^7+12t^3=t10−4t6−3t7+12t3−2t8+8t4+6t5−24t\color{Blue}-2t^8+8t^4+6t^5-24t−2t8+8t4+6t5−24t−t9+4t5+3t6−12t2\color{Blue}-t^9+4t^5+3t^6-12t^2−t9+4t5+3t6−12t2+2t7−8t3−6t4+24\color{Blue}+2t^7-8t^3-6t^4+24+2t7−8t3−6t4+24=t10−t9−2t8−t7−t6+10t5+2t4=t^{10}-t^9-2t^8-t^7-t^6+10t^5+2t^4=t10−t9−2t8−t7−t6+10t5+2t4+4t3−12t2−24t+24+4t^3-12t^2-24t+24+4t3−12t2−24t+24所以f′(x)=[t10−t9−2t8−t7−t6+10t5+2t4f'(x)=[t^{10}-t^9-2t^8-t^7-t^6+10t^5+2t^4f′(x)=[t10−t9−2t8−t7−t6+10t5+2t4+4t3−12t2−24t+24]′+4t^3-12t^2-24t+24]'+4t3−12t2−24t+24]′=10t9t′−9t8t′−16t7t′−7t6t′−6t5t′+50t4t′=10t^9t'-9t^8t'-16t^7t'-7t^6t'-6t^5t'+50t^4t'=10t9t′−9t8t′−16t7t′−7t6t′−6t5t′+50t4t′+8t3t′+12t2t′−24tt′−24t′+0+8t^3t'+12t^2t'-24tt'-24t'+0+8t3t′+12t2t′−24tt′−24t′+0=(10t9−9t8−16t7−7t6−6t5+50t4=(10t^9-9t^8-16t^7-7t^6-6t^5+50t^4=(10t9−9t8−16t7−7t6−6t5+50t4+8t3+12t2−24t−24)⋅t′+8t^3+12t^2-24t-24)\cdot t'+8t3+12t2−24t−24)⋅t′=(10e9x−9e8x−16e7x−7e6x−6e5x+50e4x=(10e^{9x}-9e^{8x}-16e^{7x}-7e^{6x}-6e^{5x}+50e^{4x}=(10e9x−9e8x−16e7x−7e6x−6e5x+50e4x+8e3x+12e2x−24ex−24)⋅(ex)′+8e^{3x}+12e^{2x}-24e^{x}-24)\cdot(e^x)'+8e3x+12e2x−24ex−24)⋅(ex)′=(10e9x−9e8x−16e7x−7e6x−6e5x+50e4x=(10e^{9x}-9e^{8x}-16e^{7x}-7e^{6x}-6e^{5x}+50e^{4x}=(10e9x−9e8x−16e7x−7e6x−6e5x+50e4x+8e3x+12e2x−24ex−24)⋅ex+8e^{3x}+12e^{2x}-24e^{x}-24)\cdot e^x+8e3x+12e2x−24ex−24)⋅ex则f′(0)=f′(x)∣x=0f'(0)=f'(x)|_{x=0}f′(0)=f′(x)∣x=0=(10e0−9e0−16e0−7e0−6e0+50e0=(10e^{0}-9e^{0}-16e^{0}-7e^{0}-6e^{0}+50e^{0}=(10e0−9e0−16e0−7e0−6e0+50e0+8e0+12e0−24e0−24)⋅e0+8e^{0}+12e^{0}-24e^{0}-24)\cdot e^0+8e0+12e0−24e0−24)⋅e0=(10−9−16−7−6+50=(10-9-16-7-6+50=(10−9−16−7−6+50+8+12−24−24)⋅1+8+12-24-24)\cdot 1+8+12−24−24)⋅1=(10−9)−(16+7)+(−6+50)\color{Blue}=(10-9)-(16+7)+(-6+50)=(10−9)−(16+7)+(−6+50)+(8+12)−(24+24)\color{Blue}+(8+12)-(24+24)+(8+12)−(24+24)=1−23+44+20−48=1-23+44+20-48=1−23+44+20−48=1+(−23+20)+(44−48)=1+(-23+20)+(44-48)=1+(−23+20)+(44−48)=1−3−4=1-3-4=1−3−4=−6=-6=−6
【花絮】
开源福利啦：供需要试玩KATEX代码的冲浪者用，用于生成本文章的KATEX代码如下：
【题目】已知$$
f(x)=(e^x-1)(e^{2x}-2)(e^{3x}-3)(e^{4x}-4)\\
$$求$f'(0).$
【解答】令$e^x=t$，则$e^{ax}=(e^x)^a=t^a$，于是
$$f(x)=(t-1)(t^2-2)(t^3-3)(t^4-4)$$$$=[(t-1)(t^2-2)]\cdot[(t^3-3)(t^4-4)]$$$$
=(t^3-2t-t^2+2)(t^7-4t^3-3t^4+12)
$$$$=\color{Blue}t^{10}-4t^6-3t^7+12t^3$$$$\color{Blue}-2t^8+8t^4+6t^5-24t$$$$\color{Blue}-t^9+4t^5+3t^6-12t^2$$$$\color{Blue}+2t^7-8t^3-6t^4+24$$$$=t^{10}-t^9-2t^8-t^7-t^6+10t^5+2t^4$$$$+4t^3-12t^2-24t+24$$所以$$f'(x)=[t^{10}-t^9-2t^8-t^7-t^6+10t^5+2t^4$$$$+4t^3-12t^2-24t+24]'$$$$=10t^9t'-9t^8t'-16t^7t'-7t^6t'-6t^5t'+50t^4t'$$$$+8t^3t'+12t^2t'-24tt'-24t'+0$$$$=(10t^9-9t^8-16t^7-7t^6-6t^5+50t^4$$$$+8t^3+12t^2-24t-24)\cdot t'$$$$=(10e^{9x}-9e^{8x}-16e^{7x}-7e^{6x}-6e^{5x}+50e^{4x}$$$$+8e^{3x}+12e^{2x}-24e^{x}-24)\cdot(e^x)'$$$$=(10e^{9x}-9e^{8x}-16e^{7x}-7e^{6x}-6e^{5x}+50e^{4x}$$$$+8e^{3x}+12e^{2x}-24e^{x}-24)\cdot e^x$$则$$f'(0)=f'(x)|_{x=0}$$$$=(10e^{0}-9e^{0}-16e^{0}-7e^{0}-6e^{0}+50e^{0}$$$$+8e^{0}+12e^{0}-24e^{0}-24)\cdot e^0$$$$=(10-9-16-7-6+50$$$$+8+12-24-24)\cdot 1$$$$\color{Blue}=(10-9)-(16+7)+(-6+50)$$$$\color{Blue}+(8+12)-(24+24)$$$$=1-23+44+20-48$$$$=1+(-23+20)+(44-48)$$$$=1-3-4$$$$=-6$$
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