对数求导法
阐述
\[已知 y=f(x) , \ \ 请使用对数求导法求 y'
\]
\(\\ \\\)
适用条件
1.幂指函数, 例如: $ \ y=x^{\sin{x}}$
2.多因子乘幂型函数, 例如:
\(\\\)
\(y = \sqrt{x^{2}(1-x^{2})\sin x}\)
\(\\\)
\(y = a^{5}b^{6}c^{7}\)
\(\\ \\\)
方法
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Step1:方程两边同时取对数
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Step2:两边同时对\(x\)求导
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Step3:计算关于\(y'\)的方程
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具体过程
\[\begin{align}
\\
已知y = f(x), 求y'
\\ \\
两边同取对数: \ln{y} = \ln{f(x)}
\\ \\
则: (\ln{y})' = \ln{f'(x)}
\\ \\
设\ln{y} = u, \ 则: \ u(y) = \ln{y},y = (y)
\\ \\
由链式法则得出:
(\ln{y})' = [u(y)]' = u' \cdot y'
\\ \\
据常见函数求导公式: u' = (\log_{e}{y})' = \frac{1}{y}
\\ \\
[u(y)]' = (\ln{y})' \cdot (y)' = \frac{1}{y} \cdot y'
\\ \\
\frac{1}{y} \cdot y' = \ln{f'(x)}
\\ \\
y' = \ln{f'(x)} \cdot y
\end{align}
\]
上述的对\(\ln{y}\)进行求导的过程,使用了关于隐函数的求导方法
例题
First
\[y=x^{x}, \ \ y'=?
\]
\[\\ \\
\]
\[两边同取对数: \ \ln{y}=\ln{x}^{x} \Rightarrow \ln{y}=x\ln{x}
\]
\[\\ \\
\]
\[(\ln{y})'=\frac{1}{y} \cdot y'
\]
\[\\ \\
\]
\[据导数乘法公式: \ (x\ln{x})'=(x)'\ln{x}+x(\ln{x})'
\]
\[\\
\]
\[\Rightarrow \ln{x}+1
\]
\[\\ \\
\]
\[\therefore \frac{1}{y} \cdot y'=\ln{x}+1
\]
\[\\
\]
\[y'=y(\ln{x}+1)
\]
\[\\ \\
\]
\[\because y=x^{x}
\]
\[\\ \\
\]
\[\therefore y'=x^{x}(\ln{x}+1)
\]
Second
\[y=(\frac{x}{1+x})^{x}, \quad y'=?
\]
\[\\ \\
\]
\[两边同时取对数: \\
\ln{y}=\ln[(\frac{x}{1+x})^{x}]\]
\[\\ \\
\]
\[\ln[(\frac{x}{1+x})^{x}] \Rightarrow x \ln{(\frac{x}{1+x})}
\]
\[\\ \\
\]
\[\Rightarrow x [\ln{x}-\ln(1+x)]
\]
\[\\ \\
\]
\[同时取导: \\ (\ln{y})'=(x [\ln{x}-\ln(1+x)])'
\]
\[\\ \\
\]
\[\frac{1}{y} \cdot y' =
(x)' \cdot [\ln{x}-\ln(1+x)] + x \cdot [\ln{x}-\ln(1+x)]'\]
\[\\ \\
\]
\[[\ln{x}-\ln(1+x)] + x \cdot (\frac{1}{x}-\frac{1}{1+x})
\]
\[\\ \\
\]
\[[\ln{x}-\ln(1+x)] + 1-\frac{x}{1+x}
\]
\[\\ \\
\]
\[y'=[\ln{(\frac{x}{1+x})} + \frac{1+x}{1+x}-\frac{x}{1+x}]
\cdot y
\]
\[\\ \\
\]
\[y'=[\ln{(\frac{x}{1+x})} + \frac{1}{1+x}] (\frac{x}{1+x})^{x}
\]
Third
\[\begin{align}
求导:y= x^{\sin{x}} , \ 其中x > 0
\\ \\
同取2边对数: \ \ln{y}= \ln{x}^{\sin{x}}
\\ \ln{y}= \sin{x}\ln{x}
\\ 据链式法则及导数乘法公式: \\
(\ln{y})'= (\sin{x}\ln{x})' \Rightarrow
(\ln{y})' \cdot y'= (\sin{x})'\ln{x}+\sin{x}(\ln{x})'
\\ \\
\frac{1}{y} \cdot y'= \cos{x}\ln{x}+\sin{x}\frac{1}{x}
\\ \\
y'= (\cos{x}\ln{x}+\sin{x}\frac{1}{x}) \div \frac{1}{y}
\\ \\
y'= x^{\sin{x}}(\cos{x}\ln{x}+\frac{\sin{x}}{x})
\end{align}
\]