复合函数求导训练
First
\[\begin{eqnarray}
y=\tan{\frac{2x}{1+x^{2}}},\quad y'=?
\\ \\
y=\tan{u}, \
u=\frac{2x}{1+x^{2}},\quad
y'=(\tan{u})'(u)'
\\ \\
(\tan{u})'=\sec^{2}{\frac{2x}{1+x^{2}}}
\\ \\
(u)'=\frac{(2x)'(1+x^{2})-(1+x^{2})2x}{(1+x^{2})^{2}}
\\ \\
(2x)'=2,\\
(1+x^{2})'=(1)'+(x^{2})'=0+2x=2x
\\ \\
\therefore (u)'=\frac{2(1+x^{2})-2x \cdot 2x}{(1+x^{2})^{2}}=\frac{2-2x^{2}}{(1+x^{2})^{2}}
\\ \\
y'=(\tan{u})'(u)'=\sec^{2}{\frac{2x}{1+x^{2}}} \cdot \frac{2-2x^{2}}{(1+x^{2})^{2}}
\end{eqnarray}
\]
Second
\[\begin{eqnarray}
y=\ln(3x^{2}+x+1),\quad
y'=?
\\ \\
y=\ln{u},\quad
u=3x^{2}+x+1,\\
\therefore y'=(\ln{u})'(u)'
\\ \\
(u)'=(3x^{2})'+(x)'+(1)'=(6x+1)+0+0
\\ \\
(\ln{u})'=\frac{1}{u\ln{e}}=\frac{1}{u}=\frac{1}{3x^{2}+x+1}
\\ \\
\therefore y'=\frac{6x+1}{3x^{2}+x+1}
\end{eqnarray}
\]
Third
\[\begin{eqnarray}
f(x)=(1+\sin{x})^{2},\quad f'(x)=?
\\ \\
设u(x)=1+\sin{x},\quad
则f(u)=u^{2}\\
f'(x)=f'[u(x)]=f'(u)\cdot u'(x)
\\ \\
f'(u)=(u^{2})'=2u=2(1+\sin{x})
\\ \\
u'(x)=(1+\sin{x})'=\cos{x}
\\ \\
\therefore f'(x)=f'[u(x)]=2(1+\sin{x})\cos{x}
\end{eqnarray}
\]
Fourth
\[\begin{eqnarray}
y=\cos{\sqrt[]{1+\ln{x}}},\quad y'=?
\\ \\
设u=\sqrt[]{1+\ln{x}}, \ 则y(u)=\cos{u}
\\ \\
设\alpha =1+\ln{x}, 则\alpha(x)=1+\ln{x}, \quad u(\alpha)=\sqrt[]{\alpha}
\\ \\
y'=y'(x)=y'\{ u[\alpha (x)] \}=y'(u)\cdot u'(\alpha )\cdot \alpha' (x)
\\ \\
y'(u) =(\cos{u})'=-\sin{u}
\\ \\
u'(\alpha )=(\sqrt[]{\alpha })'=\frac{1}{2\sqrt[]{\alpha } }
\\ \\
\alpha' (x)=(1+\ln{x})'=\frac{1}{x}
\\ \\
\frac{1}{x} \cdot \frac{1}{2\sqrt[]{\alpha}} \cdot - \sin{u}=
\frac{-\sin{\sqrt[]{1+\ln{x}}} }{2x\sqrt[]{1+\ln{x}} }
\\ \\
y'=\frac{-\sin {\sqrt[]{1+\ln{x}}} }{2x\sqrt[]{1+\ln{x}} }
\end{eqnarray}
\]
Fifth
\[\begin{eqnarray}
f'(x)=\ln\sqrt[]{1+x^{2}},\quad f'(x)=?
\\ \\
设\beta=1+x^{2}, \quad u=\sqrt[]{1+x^{2}}
\\
则\beta(x)=1+x^{2},\quad u(\beta)=\sqrt[]{1+x^{2}},\quad f(u)=\ln{u}
\\ \\
\because f(x)=f'\{ u[\beta(x)]\}\\
\therefore f'(x)=f'\{ u[\beta(x)]\}=f'(u)\cdot u'(\beta) \cdot \beta '(x)
\\ \\
f'(u)=\frac{1}{u}
\\
u'(\beta) =\frac{1}{2\sqrt[]{\beta} }
\\
\beta '(x)=2x
\\ \\
\frac{1}{u} \cdot \frac{1}{2\sqrt[]{\beta} } \cdot 2x
=\frac{1}{\sqrt[]{1+x^{2}} } \cdot \frac{1}{2\sqrt[]{1+x^{2}} } \cdot 2x
=\frac{x}{1+x^{2}}
\\ \\
\therefore f'(x)=\frac{x}{1+x^{2}}
\end{eqnarray}
\]
Sixth
本题采用了对数求导法
\[\begin{eqnarray}
y = \sqrt[4]{ \frac{(x-1)(x-2)}{(x-3)(x-4)} },\quad y'(x)=?
\\ \\
y=[\frac{(x-1)(x-2)}{(x-3)(x-4)}]^{\frac{1}{4}}
\\ \\ \\
\log{ \frac{(x-1)(x-2)}{(x-3)(x-4)} } {y}=\frac{\ln{y}}{\ln\frac{(x-1)(x-2)}{(x-3)(x-4)}}=\frac{1}{4}
\\ \\
\ln{y}=\frac{1}{4} \ln[\frac{(x-1)(x-2)}{(x-3)(x-4)}]
\\ \\
\ln{y} =\frac{1}{4} [\ln(x-1)+\ln(x-2)-\ln(x-3)-\ln(x-4)]
\\ \\
同时对两边求导: \\
\frac{d(\ln{y})}{dx} \cdot \frac{dy}{dx}=\frac{1}{4} \frac{d[\ln(x-1)+\ln(x-2)-\ln(x-3)-\ln(x-4)]}{dx}
\\ \\
\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{4} [\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}]
\\ \\
\frac{dy}{dx}=\frac{1}{4} [\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}] \cdot y
\\ \\
\frac{dy}{dx}=\frac{1}{4} [\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}] \cdot \sqrt[4]{ \frac{(x-1)(x-2)}{(x-3)(x-4)} }
\\ \\
\therefore y'(x)=\frac{1}{4} [\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}] \cdot \sqrt[4]{ \frac{(x-1)(x-2)}{(x-3)(x-4)} }
\end{eqnarray}
\]
Seventh
\[\begin{eqnarray}
(\ln |\sec x+\tan x|)^{\prime}=?
\\ \\
设: \sec x+\tan x=a
\\ \\
据链式法则,原式等于: (\ln a)^{\prime} \cdot(a)^{\prime}
\\ \\
首先: (\tan x)^{\prime}=\sec ^{2} x
\\ \\
(\sec x)^{\prime} \Rightarrow
\left( \frac{1}{\cos x} \right)^{\prime}
=\frac{\sin x}{\cos ^{2} x}
\\ \\
=\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}=\tan x \sec x
\\ \\
\therefore(a)^{\prime}=\tan x \sec x+\sec ^{2} x
\\ \\
\text { 其次: }(\ln a)^{\prime}=\frac{1}{a}
=\frac{1}{\sec x+\tan x}
\\ \\ \\
(\ln a)^{\prime} \cdot(a)^{\prime}=\frac{\tan x \sec x+\sec ^{2} x}{\sec x+\tan x}
\\ \\
=\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}
\\ \\
=\sec x
\end{eqnarray}
\]
