前置知识：函数求导简介
复合函数求导
若 y = f ( u ) y=f(u) y=f(u)， u = g ( x ) u=g(x) u=g(x)，则 d y d x = d y d u ⋅ d u d x \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx} dxdy=dudy⋅dxdu
如：
(
e
sin
x
)
′
=
e
sin
x
c
o
s
x
(e^{\sin x})'=e^{\sin x}cosx
(esinx)′=esinxcosx
[
f
(
g
(
x
)
)
]
′
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
[f(g(x))]'=f'(g(x)) \cdot g'(x)
[f(g(x))]′=f′(g(x))⋅g′(x)
证明：复合函数的导数及证明
例题
例1
y = ln sin x y=\ln \sin \sqrt x y=lnsinx,求 y ′ y' y′
解： y ′ = 1 sin x ⋅ cos x ⋅ 1 2 x = cos x 2 x sin x y'=\dfrac{1}{\sin \sqrt x}\cdot \cos \sqrt x\cdot \dfrac{1}{2\sqrt x}=\dfrac{\cos \sqrt x}{2\sqrt x\sin \sqrt x} y′=sinx1⋅cosx⋅2x1=2xsinxcosx
例2
y = ln ( x + a 2 + x 2 ) y=\ln(x+\sqrt{a^2+x^2}) y=ln(x+a2+x2)，求 y ′ y' y′
解：
y
′
=
1
x
+
a
2
+
x
2
⋅
(
1
+
1
2
a
2
+
x
2
⋅
2
x
)
=
1
x
+
a
2
+
x
2
⋅
a
2
+
x
2
+
x
a
2
+
x
2
=
1
a
2
+
x
2
y'=\dfrac{1}{x+\sqrt{a^2+x^2}}\cdot (1+\dfrac{1}{2\sqrt{a^2+x^2}}\cdot2x)=\dfrac{1}{x+\sqrt{a^2+x^2}}\cdot\dfrac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}}=\dfrac{1}{\sqrt{a^2+x^2}}
y′=x+a2+x21⋅(1+2a2+x21⋅2x)=x+a2+x21⋅a2+x2a2+x2+x=a2+x21
例3
y
=
f
(
ln
x
)
e
f
(
x
)
y=f(\ln x)e^{f(x)}
y=f(lnx)ef(x)，求
y
′
y'
y′
解：
y
′
=
f
′
(
ln
x
)
⋅
1
x
⋅
e
f
(
x
)
+
f
(
ln
x
)
⋅
e
f
(
x
)
⋅
f
′
(
x
)
=
f
′
(
ln
x
)
e
f
(
x
)
x
+
f
(
ln
x
)
⋅
e
f
(
x
)
⋅
f
′
(
x
)
y'=f'(\ln x)\cdot \dfrac 1x \cdot e^{f(x)}+f(\ln x)\cdot e^{f(x)}\cdot f'(x)=\dfrac{f'(\ln x)e^{f(x)}}{x}+f(\ln x)\cdot e^{f(x)}\cdot f'(x)
y′=f′(lnx)⋅x1⋅ef(x)+f(lnx)⋅ef(x)⋅f′(x)=xf′(lnx)ef(x)+f(lnx)⋅ef(x)⋅f′(x)
扫一扫
2万+
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
