Pembuktian turunan Ln x

Posted on December 6, 2015 by Nursatria

Yang belajar turunan pastinya tahu bahwa turunan dari ln x adalah 1/x

{\displaystyle \frac{d}{dx}\ln x=\frac{1}{x}}

Sekarang mari kita buktikan

Berdasarkan definsi turunan

{\displaystyle \lim_{h\rightarrow0}\frac{\ln\left(x+h\right)-\ln x}{h}}

Gunakan sifat-sifat logaritma

{\displaystyle \lim_{h\rightarrow0}\frac{\ln\left(\frac{x+h}{x}\right)}{h}=\lim_{h\rightarrow0}\ln\left(\frac{x+h}{x}\right)^{\frac{1}{h}}}

Gunakan aljabar

{\displaystyle \lim_{h\rightarrow0}\ln\left(1+\frac{h}{x}\right)^{\frac{1}{h}}}

Diketahui {\displaystyle \frac{1}{h}=\frac{x}{h}\cdot\frac{1}{x}}

{\displaystyle \lim_{h\rightarrow0}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}\cdot\frac{1}{x}}=\lim_{h\rightarrow0}\ln\left(\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right)^{\frac{1}{x}}}

{\displaystyle \lim_{h\rightarrow0}\frac{1}{x}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}}=\frac{1}{x}\lim_{h\rightarrow0}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}}}

Selanjutnya subtitusi n=x/h , jika h\rightarrow0 maka n\rightarrow\infty

{\displaystyle \frac{1}{x}\lim_{n\rightarrow\infty}\ln\left(1+\frac{1}{n}\right)^{n}}

Sekarang, perhatikan \left(1+1/n\right)^{n} untuk n\rightarrow\infty merupakan definisi dari e. So.. dapat disimpulkan

{\displaystyle \frac{d}{dx}\ln x=\frac{1}{x}\lim_{n\rightarrow\infty}\ln e=\frac{1}{x}}

∴

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About Nursatria

Seorang Alumnus Matematika UGM, dengan ilmu yang didapat ketika kuliah (Padahal sering bolos kuliah :p ), saya menyebarkan virus matematika
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