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I'm currently working on trying to solve for a derivative using the formal definition $\lim_{h\to0}\frac{f(x_0 +h)-f(x_0)}{h}$ I'm solving for $\frac{\text{d}y}{\text{d}x}$for $y=\frac{4}{x}$. Using the rules of derivatives I know the derivative is $y=\frac{-4}{x^2}$, but I'm confused on how to get this from the definition. After plugging into the formula and simplifying to get rid of the complex fraction I got $\lim_{h\to0} \frac{\frac{-4h}{x^2+hx}}{h}.$ How can I simplify that expression to $\frac{-4}{x^2}$? Thanks for any advice you may have.

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  • $\begingroup$ As you mentioned, you have $$\lim_{h \rightarrow 0} \frac{\frac{-4h}{x^2+hx}}{h} = \lim_{h \rightarrow 0} \frac{-4h}{h(x^2+2h)}=$$ $$\frac{-4}{(x^2+2(h=0))}=\frac{-4}{x^2}$$ $\endgroup$ – FundThmCalculus Dec 12 '16 at 20:59
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$$\frac{f(x+h)-f(x)}{h}=\frac{4}{h}(\frac{1}{x+h}-\frac{1}{x})$$

$$=\frac{4}{h}\frac{-h}{x(x+h)}$$

$$=\frac{-4}{x(x+h)}$$

and the limit when $h\to 0\;$ is $$\frac{-4}{x^2}=f'(x).$$

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