3.1 Defining the derivative  (Page 2/10)

In [link] (a) we see that, as the values of $x$ approach $a,$ the slopes of the secant lines provide better estimates of the rate of change of the function at $a.$ Furthermore, the secant lines themselves approach the tangent line to the function at $a,$ which represents the limit of the secant lines. Similarly, [link] (b) shows that as the values of $h$ get closer to $0,$ the secant lines also approach the tangent line. The slope of the tangent line at $a$ is the rate of change of the function at $a,$ as shown in [link] (c).

In [link] we show the graph of $f\left(x\right)=\sqrt{x}$ and its tangent line at $(1,1)$ in a series of tighter intervals about $x=1.$ As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of $x$ close to $1.$ In fact, the graph of $f(x)$ itself appears to be locally linear in the immediate vicinity of $x=1.$

Formally we may define the tangent line to the graph of a function as follows.

Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines.

Finding a tangent line

Find the equation of the line tangent to the graph of $f\left(x\right)=x^2$ at $x=3.$

First find the slope of the tangent line. In this example, use [link] .

$\begin{array}{ccccc}\hfill m_{\text{tan}}& =\underset{x\to 3}{\text{lim}}\frac{f\left(x\right)-f\left(3\right)}{x-3}\hfill & & & \text{Apply the definition.}\hfill \\ & =\underset{x\to 3}{\text{lim}}\frac{x^2-9}{x-3}\hfill & & & \text{Substitute}f\left(x\right)=x^2\text{and}f\left(3\right)=9.\hfill \\ & =\underset{x\to 3}{\text{lim}}\frac{\left(x-3\right)\left(x+3\right)}{x-3}=\underset{x\to 3}{\text{lim}}\left(x+3\right)=6\hfill & & & \text{Factor the numerator to evaluate the limit.}\hfill \end{array}$

Next, find a point on the tangent line. Since the line is tangent to the graph of $f(x)$ at $x=3,$ it passes through the point $\left(3,f\left(3\right)\right).$ We have $f\left(3\right)=9,$ so the tangent line passes through the point $\left(3,9\right).$

Using the point-slope equation of the line with the slope $m=6$ and the point $\left(3,9\right),$ we obtain the line $y-9=6\left(x-3\right).$ Simplifying, we have $y=6x-9.$ The graph of $f\left(x\right)=x^2$ and its tangent line at $3$ are shown in [link] .

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