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Limit Definition of the Derivative

The geometric meaning of the derivative

\begin{displaymath}f'(x)=\frac{df(x)}{dx}\end{displaymath}

is the slope of the line tangent to y = f(x) at x.

Let's look for this slope at P:

The secant line through P and Q has slope

\begin{displaymath}\frac{f(x+\Delta x)-f(x)}{(x+\Delta x)-x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}.\end{displaymath}

Figure 1

We can approximate the tangent line through P by moving Q towards P, decreasing Δx. In the limit as lim delta x to 0, we get the tangent line through P with slope

\begin{displaymath}\lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}.\end{displaymath}

We define

\begin{displaymath}f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}^\mathrm{*}.\end{displaymath}

* If the limit as lim 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.

Example

For $f(x)=x^2$,

\begin{eqnarray*} f'(x)&=&\lim_{\Delta x\to 0} \frac{(x+\Delta x)^2-x^2}{\Delta ... ...2}{\Delta x}\\ &=&\lim_{\Delta x\to 0} (2x+\Delta x)\\ &=& 2x \end{eqnarray*}

as expected.

Example

For $\displaystyle f(x)=\frac{1}{x}$

\begin{eqnarray*} f'(x)&=&\lim_{\Delta x\to 0}\frac{\frac{1}{x+\Delta x}-\frac{1... ...{\Delta x\to 0} \frac{-1}{(x+\Delta x)(x)}\\ &=& -\frac{1}{x^2} \end{eqnarray*}

again as expected.

Notes

The limit definition of the derivative is used to prove many well-known results, including the following:

Key Concept

We define

$f'(x) = \displaystyle\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$.

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