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如何应用对数求导法求函数的导数？

$\frac{1}{y}y’=\frac{f(x)}{f'(x)},$

1，函数是幂指函数 $$y=h(x)^{g(x)}$$ 的情形。例如
$y=\sin x ^{\ln x}$

$\ln y=\ln(\sin x ^{\ln x})。$

$\ln y= \ln x \ln(\sin x).$

\begin{align*}\frac{1}{y}y’&=\frac{1}{x}\ln(\sin x)+\ln x \frac{\cos x}{\sin x}\\ &=\frac{\ln(\sin x)}{x}+\ln x\tan x \end{align*}.

\begin{align*} y’&=y\left(\frac{\ln(\sin x)}{x}+\ln x\tan x\right)\\ &=\sin x ^{\ln x}\left(\frac{\ln(\sin x)}{x}+\ln x\tan x\right) \end{align*}

2，函数混合了多重乘、除法及根式，例如
$y=\frac{\sqrt[3]{7x^2+1}\cdot \sqrt[5]{2x-3}}{\sqrt{x^2+5}\cdot \sqrt[4]{3x-2}}.$

$\ln y =\ln \left(\frac{\sqrt[3]{7x^2+1}\cdot \sqrt[5]{2x-3}}{\sqrt{x^2+5}\cdot \sqrt[4]{3x-2}}\right)$

$\ln y=\frac{1}{3}\ln(7x^2+1)+\frac{1}{5}\ln(2x-3)-\frac{1}{2}\ln(x^2+5)-\frac{1}{4}\ln(3x-2)$

$\frac{y’}{y}=\frac{1}{3}\frac{14x}{7x^2+1}+\frac{1}{5}\frac{2}{2x-3}-\frac{1}{2}\frac{2x}{x^2+5}-\frac{1}{4}\frac{3}{3x-2}$

$y’=\frac{\sqrt[3]{7x^2+1}\cdot \sqrt[5]{2x-3}}{\sqrt{x^2+5}\cdot \sqrt[4]{3x-2}}\left(\frac{1}{3}\frac{14x}{7x^2+1}+\frac{1}{5}\frac{2}{2x-3}-\frac{1}{2}\frac{2x}{x^2+5}-\frac{1}{4}\frac{3}{3x-2 }\right)$