|
求
f
(
x
)
=
s
i
n
x
f(x) = sin x
f(x)=sinx的导数
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
x
h
f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}
f′(x)=h→0limhf(x+h)−f(x)=h→0limhsin(x+h)−sinx 下一步需要用到和差角公式
sin
(
x
+
y
)
=
sin
x
cos
y
+
sin
y
cos
x
\sin(x+y) = \sin x \cos y+\sin y\cos x
sin(x+y)=sinxcosy+sinycosx 所以
f
′
(
x
)
=
lim
h
→
0
sin
x
cos
h
+
sin
h
cos
x
−
sin
x
h
f'(x) = \lim_{h\to0}\frac{\sin x \cos h+\sin h\cos x-\sin x}{h}
f′(x)=h→0limhsinxcosh+sinhcosx−sinx
f
′
(
x
)
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
+
sin
h
cos
x
h
f'(x) = \lim_{h\to0}\frac{\sin x (\cos h-1)+\sin h\cos x}{h}
f′(x)=h→0limhsinx(cosh−1)+sinhcosx 我们将式子拆解成两个部分，分别进行处理
f
′
(
x
)
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
h
+
lim
h
→
0
sin
h
cos
x
h
f'(x) = \lim_{h\to0}\frac{\sin x (\cos h-1)}{h}+\lim_{h\to0}\frac{\sin h\cos x}{h}
f′(x)=h→0limhsinx(cosh−1)+h→0limhsinhcosx
对于
f
′
(
x
)
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
h
f'(x) = \lim_{h\to0}\frac{\sin x (\cos h-1)}{h}
f′(x)=h→0limhsinx(cosh−1)部分，需要用到半倍角公式
sin
2
x
2
=
1
−
cos
x
2
\sin^2 \frac{x}{2} = \frac{1-\cos x}{2}
sin22x=21−cosx 代入可得
f
′
(
x
)
=
lim
h
→
0
−
2
sin
x
sin
2
h
2
h
f'(x) = \lim_{h\to0}-2\frac{\sin x \sin^2 \frac{h}{2}}{h}
f′(x)=h→0lim−2hsinxsin22h对于
f
′
(
x
)
=
lim
h
→
0
sin
h
cos
x
h
f'(x) = \lim_{h\to0}\frac{\sin h\cos x}{h}
f′(x)=h→0limhsinhcosx部分，需要用到二倍角公式
sin
2
x
=
2
sin
x
cos
x
\sin 2x =2 \sin x \cos x
sin2x=2sinxcosx 代入可得
f
′
(
x
)
=
lim
h
→
0
2
sin
h
2
cos
h
2
cos
x
h
f'(x) = \lim_{h\to0}\frac{2 \sin \frac{h}{2} \cos \frac{h}{2} \cos x}{h}
f′(x)=h→0limh2sin2hcos2hcosx
将这两个部分代入可得
f
′
(
x
)
=
lim
h
→
0
2
sin
h
2
cos
h
2
cos
x
−
2
sin
x
sin
2
h
2
h
f'(x) = \lim_{h\to0}\frac{2 \sin \frac{h}{2} \cos \frac{h}{2} \cos x-2\sin x \sin^2 \frac{h}{2}}{h}
f′(x)=h→0limh2sin2hcos2hcosx−2sinxsin22h 提取
2
sin
h
2
2\sin \frac{h}{2}
2sin2h
f
′
(
x
)
=
lim
h
→
0
2
sin
h
2
(
cos
h
2
cos
x
−
sin
x
sin
h
2
)
h
f'(x) = \lim_{h\to0}\frac{2 \sin \frac{h}{2} (\cos \frac{h}{2} \cos x-\sin x \sin \frac{h}{2})}{h}
f′(x)=h→0limh2sin2h(cos2hcosx−sinxsin2h) 下一步需要用到和差角公式
cos
(
x
+
y
)
=
cos
x
cos
y
−
sin
x
sin
y
\cos (x+y) = \cos x \cos y -\sin x \sin y
cos(x+y)=cosxcosy−sinxsiny 代入可得
f
′
(
x
)
=
lim
h
→
0
2
sin
h
2
(
cos
(
x
+
h
2
)
)
h
f'(x) = \lim_{h\to0}\frac{2 \sin \frac{h}{2} (\cos(x+\frac{h}{2}))}{h}
f′(x)=h→0limh2sin2h(cos(x+2h))
f
′
(
x
)
=
lim
h
→
0
cos
(
x
+
h
2
)
⋅
lim
h
→
0
sin
h
2
h
2
=
cos
x
⋅
1
=
cos
x
f'(x) = \lim_{h\to0}\cos (x+\frac{h}{2}) · \lim_{h\to0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}=\cos x · 1 = \cos x
f′(x)=h→0limcos(x+2h)⋅h→0lim2hsin2h=cosx⋅1=cosx
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