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Once we know the most basic differentiation formulas and rules, we compute new derivatives using what we already know. We rarely think back to where the basic formulas and rules originated. The geometric meaning of the derivative \[f'(x)=\frac{df(x)}{dx}\] is the slope of the line tangent to $y=f(x)$ at $x$.
The secant line through $P$ and $Q$ has slope \[\frac{f(x+\Delta x)-f(x)}{(x+\Delta x)-x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}.\] We can approximate the tangent line through $P$ by moving $Q$ towards $P$, decreasing $\Delta x$. In the limit as $\Delta x\to 0$, we get the tangent line through $P$ with slope \[\lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}.\] We define \[f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)^{\small\textrm{*}}}{\Delta x}.\] $^*$ If the limit as $\Delta x \to 0$ at a particular point does not exist, $f'(x)$ is undefined at that point. We derive all the basic differentiation formulas using this definition.
ExampleFor $f(x)=x^2$, \begin{eqnarray*} f'(x)&=&\lim_{\Delta x\to 0} \frac{(x+\Delta x)^2-x^2}{\Delta x}\\ &=& \lim_{\Delta x\to 0} \frac{(x^2+2(\Delta x)x+\Delta x^2)-x^2}{\Delta x}\\ &=&\lim_{\Delta x\to 0} \frac{2(\Delta x)x+\Delta x^2}{\Delta x}\\ &=&\lim_{\Delta x\to 0} (2x+\Delta x)\\ &=& 2x \end{eqnarray*} as expected.
ExampleFor $\displaystyle f(x)=\frac{1}{x}$ \begin{eqnarray*} f'(x)&=&\lim_{\Delta x\to 0}\frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x} \\ &=&\lim_{\Delta x\to 0}\frac{\frac{x-(x+\Delta x)}{(x+\Delta x)(x)}}{\Delta x} \\ &=&\lim_{\Delta x\to 0}\frac{\frac{-\Delta x}{(x+\Delta x)(x)}}{\Delta x} \\ &=&\lim_{\Delta x\to 0} \frac{-1}{(x+\Delta x)(x)}\\ &=& -\frac{1}{x^2} \end{eqnarray*} again as expected.
The limit definition of the derivative is used to prove many well-known results, including the following:
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