[tex]\dfrac{dy}{dx}=\bf\dfrac{1}{x\,\ln|x|}[/tex]
Pembahasan
Turunan
Yang perlu diingat:
[tex]\begin{aligned}\bullet\quad&\frac{d\bigl(f(u)\bigr)}{dx}=\frac{df}{dx}\cdot\frac{du}{dx}\\\bullet\quad&\frac{d}{dx}\bigl(\ln(x)\bigr)=\frac{1}{x}\\\bullet\quad&\frac{d}{dx}\bigl(|x|\bigr)=\frac{x}{|x|}\\\end{aligned}[/tex]
dan tentunya identitas logaritma.
Penyelesaian
[tex]\begin{aligned}y&=\ln\left|\ln\left|\frac{1}{x}\right|\right|\\\\&\quad\left[\ \begin{aligned}\left|\ln\left|\frac{1}{x}\right|\right|&=\left|\ln\left(|x|\right)^{-1}\right|\\&=\Bigl|-\ln|x|\Bigr|\\&=\ln|x|\end{aligned}\right.\\\\\therefore\ y&=\ln\left(\ln|x|\right)\\\end{aligned}[/tex]
[tex]\begin{aligned}\frac{dy}{dx}&=\frac{d}{dx}\Bigl(\ln\bigl(\ln|x|\bigr)\Bigr)\\&=\frac{d}{dx}\bigl(\ln(u)\bigr)\,,\ u=\ln|x|\\&=\frac{1}{u}\cdot\frac{du}{dx}\\&=\frac{1}{\ln|x|}\cdot\frac{d}{dx}\bigl(\ln|x|\bigr)\\&=\frac{1}{\ln|x|}\cdot\frac{1}{|x|}\cdot\frac{d}{dx}\bigl(|x|\bigr)\\&=\frac{1}{\ln|x|}\cdot\frac{1}{|x|}\cdot\frac{x}{|x|}\\&=\frac{1}{\ln|x|}\cdot\frac{x}{x^2}\\&=\frac{1}{\ln|x|}\cdot\frac{1}{x}\\&=\boxed{\ \bf\frac{1}{x\,\ln|x|}\ }\end{aligned}[/tex]
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