I am struggling to find the derivative of $f(x)=\frac1x+2$ (and/or $f(x)=\frac1{x+2}$) using the definition of the derivative. thank you
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Do you know the definition? – Chris Eagle Jul 22 '13 at 13:14
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5Is that meant to be $\frac1x + 2$ or $\frac{1}{x+2}$? – Daniel Fischer Jul 22 '13 at 13:15
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@ Daniel Fischer , the second part you wrote is what I am asking, 1/x+2 – pashy Jul 22 '13 at 13:28
1 Answers
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- First case
$$f'(x):=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\frac{1}{h}\left(\frac{1}{x+h}-\frac{1}{x} \right)=\lim_{h\rightarrow 0}\frac{1}{h}\left( \frac{-h}{x(x+h)}\right)=\lim_{h\rightarrow 0} \frac{-1}{x(x+h)}=-\frac{1}{x^2}$$
- Second case
$$f'(x):=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\frac{1}{h}\left(\frac{1}{x+2+h}-\frac{1}{x+2} \right)=\lim_{h\rightarrow 0}\frac{1}{h}\left( \frac{-h}{(x+2)(x+2+h)}\right)=\lim_{h\rightarrow 0} \frac{-1}{(x+2)(x+2+h)}=-\frac{1}{(x+2)^2}$$
Avitus
- 14,018
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For the second function the strategy is similar: once you write the difference in the definition of the derivative explicitly you arrive at a single fraction $\frac{-1}{(x+2)(x+2+h)}$, whose limit for $h\rightarrow 0$ is $\frac{-1}{(x+2)^2}$. – Avitus Jul 22 '13 at 13:30
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