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Jak wyprowadzić wzór na pochodną takiej funkcji \(\frac{1}{x}\) ?

\(\lim_{h\to 0 } \frac{f(x+h)-f(x)}{h}= \lim_{h\to 0 } \frac{ \frac{1}{x+h}- \frac{1}{x} }{h}= \lim_{h\to 0 } \frac{ \frac{x-x-h}{x(x+h)} }{h}= \lim_{h\to 0} \frac{ \frac{-h}{x^2+hx} }{h}=\\
= \lim_{h\to 0 } \frac{-h}{x^2+hx} \cdot \frac{1}{h}= \lim_{h\to 0} \frac{-1}{x^2+hx}= \frac{-1}{x^2}\)

\(\lim_{\Delta x\to0}\ \frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{x\to0}\ \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}=\lim_{\Delta x\to0}\ \frac{\frac{x-x-\Delta x}{x\cdot(x+\Delta x)}}{\Delta x}=\\=\lim_{\Delta x\to0}\ \frac{-\Delta x}{\Delta x \cdot x(x+\Delta x)}=\lim_{\Delta x\to0}\ \frac{-1}{x(x+\Delta x)}=-\frac{1}{x^2}\)