【题目】已知 f ( x ) = ( e x − 1 ) ( e 2 x − 2 ) ( e 3 x − 3 ) ( e 4 x − 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
e
x
=
t
e^x=t
ex=t，则
e
a
x
=
(
e
x
)
a
=
t
a
e^{ax}=(e^x)^a=t^a
eax=(ex)a=ta，于是
f
(
x
)
=
(
t
−
1
)
(
t
2
−
2
)
(
t
3
−
3
)
(
t
4
−
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
)
(
t
2
−
2
)
]
⋅
[
(
t
3
−
3
)
(
t
4
−
4
)
]
=[(t-1)(t^2-2)]\cdot[(t^3-3)(t^4-4)]
=[(t−1)(t2−2)]⋅[(t3−3)(t4−4)]
=
(
t
3
−
2
t
−
t
2
+
2
)
(
t
7
−
4
t
3
−
3
t
4
+
12
)
=(t^3-2t-t^2+2)(t^7-4t^3-3t^4+12)
=(t3−2t−t2+2)(t7−4t3−3t4+12)
=
t
10
−
4
t
6
−
3
t
7
+
12
t
3
=\color{Blue}t^{10}-4t^6-3t^7+12t^3
=t10−4t6−3t7+12t3
−
2
t
8
+
8
t
4
+
6
t
5
−
24
t
\color{Blue}-2t^8+8t^4+6t^5-24t
−2t8+8t4+6t5−24t
−
t
9
+
4
t
5
+
3
t
6
−
12
t
2
\color{Blue}-t^9+4t^5+3t^6-12t^2
−t9+4t5+3t6−12t2
+
2
t
7
−
8
t
3
−
6
t
4
+
24
\color{Blue}+2t^7-8t^3-6t^4+24
+2t7−8t3−6t4+24
=
t
10
−
t
9
−
2
t
8
−
t
7
−
t
6
+
10
t
5
+
2
t
4
=t^{10}-t^9-2t^8-t^7-t^6+10t^5+2t^4
=t10−t9−2t8−t7−t6+10t5+2t4
+
4
t
3
−
12
t
2
−
24
t
+
24
+4t^3-12t^2-24t+24
+4t3−12t2−24t+24所以
f
′
(
x
)
=
[
t
10
−
t
9
−
2
t
8
−
t
7
−
t
6
+
10
t
5
+
2
t
4
f'(x)=[t^{10}-t^9-2t^8-t^7-t^6+10t^5+2t^4
f′(x)=[t10−t9−2t8−t7−t6+10t5+2t4
+
4
t
3
−
12
t
2
−
24
t
+
24
]
′
+4t^3-12t^2-24t+24]'
+4t3−12t2−24t+24]′
=
10
t
9
t
′
−
9
t
8
t
′
−
16
t
7
t
′
−
7
t
6
t
′
−
6
t
5
t
′
+
50
t
4
t
′
=10t^9t'-9t^8t'-16t^7t'-7t^6t'-6t^5t'+50t^4t'
=10t9t′−9t8t′−16t7t′−7t6t′−6t5t′+50t4t′
+
8
t
3
t
′
+
12
t
2
t
′
−
24
t
t
′
−
24
t
′
+
0
+8t^3t'+12t^2t'-24tt'-24t'+0
+8t3t′+12t2t′−24tt′−24t′+0
=
(
10
t
9
−
9
t
8
−
16
t
7
−
7
t
6
−
6
t
5
+
50
t
4
=(10t^9-9t^8-16t^7-7t^6-6t^5+50t^4
=(10t9−9t8−16t7−7t6−6t5+50t4
+
8
t
3
+
12
t
2
−
24
t
−
24
)
⋅
t
′
+8t^3+12t^2-24t-24)\cdot t'
+8t3+12t2−24t−24)⋅t′
=
(
10
e
9
x
−
9
e
8
x
−
16
e
7
x
−
7
e
6
x
−
6
e
5
x
+
50
e
4
x
=(10e^{9x}-9e^{8x}-16e^{7x}-7e^{6x}-6e^{5x}+50e^{4x}
=(10e9x−9e8x−16e7x−7e6x−6e5x+50e4x
+
8
e
3
x
+
12
e
2
x
−
24
e
x
−
24
)
⋅
(
e
x
)
′
+8e^{3x}+12e^{2x}-24e^{x}-24)\cdot(e^x)'
+8e3x+12e2x−24ex−24)⋅(ex)′
=
(
10
e
9
x
−
9
e
8
x
−
16
e
7
x
−
7
e
6
x
−
6
e
5
x
+
50
e
4
x
=(10e^{9x}-9e^{8x}-16e^{7x}-7e^{6x}-6e^{5x}+50e^{4x}
=(10e9x−9e8x−16e7x−7e6x−6e5x+50e4x
+
8
e
3
x
+
12
e
2
x
−
24
e
x
−
24
)
⋅
e
x
+8e^{3x}+12e^{2x}-24e^{x}-24)\cdot e^x
+8e3x+12e2x−24ex−24)⋅ex则
f
′
(
0
)
=
f
′
(
x
)
∣
x
=
0
f'(0)=f'(x)|_{x=0}
f′(0)=f′(x)∣x=0
=
(
10
e
0
−
9
e
0
−
16
e
0
−
7
e
0
−
6
e
0
+
50
e
0
=(10e^{0}-9e^{0}-16e^{0}-7e^{0}-6e^{0}+50e^{0}
=(10e0−9e0−16e0−7e0−6e0+50e0
+
8
e
0
+
12
e
0
−
24
e
0
−
24
)
⋅
e
0
+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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