What Does the Definition of the Derivative Mean?

The derivative of a function $f (x)$ at $x = x_{1}$ is a number that tells us how much the graph of the function $f (x)$ is increasing (or decreasing) when the $x$ -coordinate is $x_{1}$ . The derivative at $x = x_{1}$ is written as $f^{'} (x_{1})$ . The relationships between the derivative, the instantaneous rate of change, and the slope of the tangent to the graph $f (x)$ is described below.

Theory

The Derivative, the Instantaneous Rate of Change, and the Slope of the Tangent

\begin{array}{l} f^{'} (x_{1}) \\ = the instantaneous rate of \\ change at x_{1} \\ = the slope of the tangent to \\ f (x) in the point (x_{1}, f (x_{1})) \end{array}

\begin{array}{l} f^{'} (x_{1}) & = the instantaneous rate of change at x_{1} \\ = the slope of the tangent to f (x) in the point (x_{1}, f (x_{1})) . \end{array}

Before you read the explanation below, you need to know what $Δ x$ and $x + Δ x$ means ( $Δ x$ is read as “delta $x$ ”).

$Δ x$ means “change in the value of $x$ ”, so it represents the distance between two $x$ -values.
$x + Δ x$ means a distance $Δ x$ from $x$ .

Explanation of the Derivative

Look closely at the figures below while you read the text.

Figure 1: The graph of the function in blue and the secant in pink. The points of intersection are

(x, f (x))

(x + Δ x, f (x + Δ x))

You have the function $f (x)$ (the blue graph), and you have drawn a secant (the pink line) between the points $(x, f (x))$ and $(x + Δ x, f (x + Δ x))$ . As you’ve now learned, the slope of the pink line is the average rate of change of the function between these points. In addition you’ve learned that if these points lie close together, then the average rate of change approaches the instantaneous rate of change.

The derivative of a function 2

Figure 2: The same situation as in the previous figure. However, now the point

(x + Δ x, f (x + Δ x))

is closer to

(x, f (x))

, and thus

Δ x

is smaller.

By reducing $Δ x$ , the rightmost point of intersection $(x + Δ x, f (x + Δ x))$ approaches the leftmost point of intersection $(x, f (x))$ . When this happens, the distance between $x$ and $x + Δ x$ , as well as the distance between $f (x)$ and $f (x + Δ x)$ , becomes smaller. The slope of the secant is gradually approaching the slope of the tangent of the function at $(x, f (x))$ .

Figure 3: The same situation as in the two previous figures. However, this time you’ve let

Δ x

approach zero, and the pink line is now tangent to

f (x)

As the distance between the two points of intersection approaches zero, the pink line will touch

f (x)

(x, f (x))

. The slope of the tangent is equal to the instantaneous rate of change at that point.

The value of $x$ was arbitrary. This means that if you formulate your approach mathematically, you get a new function for the slope of $f (x)$ for all values of $x$ . This is precisely what the derivative $f^{'} (x)$ of a function is.

By using what you’ve learned regarding instantaneous rate of change and limits, you get the following definition of the derivative:

Theory

Definition of the Derivative

f^{'} (x) = \lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}

Example 1

Given $f (x) = x^{2}$ , you can differentiate $f (x)$ by using the definition:

\begin{array}{l} f^{'} (x) & = \lim_{Δ x \to 0} \frac{{(x + Δ x)}^{2} - x^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{(x + Δ x) (x + Δ x) - x^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{x^{2} + 2 x Δ x + {(Δ x)}^{2} - x^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{2 x Δ x + {(Δ x)}^{2}}{Δ x} \\ = \lim_{Δ x \to 0} \frac{Δ x (2 x + Δ x)}{Δ x} \\ = \lim_{Δ x \to 0} \frac{Δx (2 x + Δ x)}{Δx} \\ = \lim_{Δ x \to 0} (2 x + Δ x) \\ = 2 x + 0 \\ = 2 x \end{array}

It’s pretty rare to differentiate a function by using the definition of the derivative. Most of the time, you use the rules for derivation.

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