The Derivative Function

Section 2.1 The Derivative Function

Recall that previously, we found the instantaneous rate of change of a function, which is the slope of the tangent line to the graph of the function, using a limit of the form,

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\begin{equation*} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \end{equation*}

This limit is very important in calculus, so it is given a name and a concise notation, called the derivative.

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Definition 2.1.1. Derivative at a Point.

The derivative of $f$ at $a\text{,}$ denoted by $f'(a)$ (read as “$f$ prime of $a$”), is given by,

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\begin{equation*} \boxed{f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}} \end{equation*}

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Example 2.1.2. Derivative at a Specific Point.

Find the derivative of $f(x) = x^2 + 2$ at $x = 3\text{.}$

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\begin{align*} f'(3) \amp= \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}\\ \amp= \lim_{h \to 0} \frac{(3 + h)^2 + 2 - (3^2 + 2)}{h}\\ \amp= \lim_{h \to 0} \frac{(9 + 6h + h^2) + 2 - 11}{h} \amp\amp \text{expanding and simplifying}\\ \amp= \lim_{h \to 0} \frac{6h + h^2}{h} \amp\amp \text{combining like terms}\\ \amp= \lim_{h \to 0} (6 + h) \amp\amp \text{cancelling a common factor of $h$}\\ f'(3) \amp= 6 \amp\amp \text{plugging in $h=0$} \end{align*}

Therefore, $f'(3) = 6\text{.}$

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If we want to find the slope of the tangent line at many different values of $x\text{,}$ then this way of using the definition for each value of $x$ is not very efficient.

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Instead, we can calculate the derivative at an arbitrary value of $x\text{,}$ which will give us a formula for the slope of the tangent at any value of $x\text{.}$

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Example 2.1.3. Squaring Function.

Find the derivative of $f(x) = x^2$ at an arbitrary value of $x\text{.}$ To do this, instead of plugging in a specific value for $x\text{,}$ we will just leave it as $x\text{,}$

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\begin{align*} f'(x) \amp= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\\ \amp= \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h}\\ \amp= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h} \amp\amp \text{expanding}\\ \amp= \lim_{h \to 0} \frac{2xh + h^2}{h} \amp\amp \text{combining like terms}\\ \amp= \lim_{h \to 0} (2x + h) \amp\amp \text{cancelling a common factor of $h$}\\ f'(x) \amp= 2x \amp\amp \text{plugging in $h=0$} \end{align*}

Therefore, $f'(x)=2x\text{.}$ This means that the slope of the tangent line to the graph of $f$ at any value of $x$ is $2x\text{.}$ For example, at $x=3\text{,}$ the slope of the tangent line is $f'(3) = 2 \cdot 3 = 6\text{,}$ which is what we found in the previous example.

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This allows us to create a table of values for the slope of the tangent line at different values of $x\text{,}$

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$x$	$f'(x)$
$-2$	$-4$
$-1$	$-2$
$0$	$0$
$1$	$2$
$2$	$4$

This gives the slope at each of these values of $x\text{.}$ This means that the derivative is itself a function, which takes in a value of $x\text{,}$ and outputs the slope of the tangent line at that value of $x\text{.}$ Here is a graph: Graph of $f(x)=x^2$ and its derivative. At each value of $x\text{,}$ the slope of the tangent line to the graph of $f$ is given by the $y$-value of $f'(x)=2x\text{.}$

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In general, the derivative of a function $f$ is not just a number, but is itself a function.

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Subsection 2.1.1 The Derivative as a Function

Definition 2.1.4. Derivative Function.

Let $f$ be a function. The derivative of $f\text{,}$ denoted by $f'\text{,}$ is another function, defined by,

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\begin{equation*} \boxed{f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}} \end{equation*}

The derivative “$f'$” is read as “$f$ prime” and “$f'(x)$” as “$f$ prime of $x$”.
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The action of finding the derivative of a function is often called “taking the derivative” or “differentiating”.
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Example 2.1.5. Reciprocal Function.

Consider the derivative of $f(x) = \frac{1}{x}\text{,}$

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\begin{align*} f'(x) \amp= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\\ \amp= \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} \end{align*}

To simplify this difference quotient, we first simplify the complex fraction, by clearing denominators, by multiplying numerator and denominator by $x(x+h)\text{,}$

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\begin{align*} \amp= \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} \cdot \frac{x(x + h)}{x(x + h)}\\ \amp= \lim_{h \to 0} \frac{x - (x + h)}{hx(x + h)}\\ \amp= \lim_{h \to 0} -\frac{h}{hx(x + h)} \amp\amp \text{simplifying the numerator}\\ \amp= \lim_{h \to 0} -\frac{1}{x(x + h)} \amp\amp \text{cancelling the common factor of $h$}\\ \amp= -\frac{1}{x^2} \end{align*}

Therefore, $f'(x) = -\frac{1}{x^2}\text{.}$ Observe the relationship between $f(x)$ and its derivative: Graph of $f(x)=\frac{1}{x}$ and its derivative. Notice that:

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The slope of the tangent of $f$ is negative for all values of $x\text{,}$ which corresponds to the derivative being negative (or, $f'(x) \lt 0$).
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As $x$ gets larger, the slope of the tangent line gets closer to $0\text{,}$ which corresponds to the derivative getting closer to $0$ (or, $f'(x) \to 0$ as $x \to \infty$).
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Example 2.1.6. Square Root Function.

Consider the derivative of $f(x) = \sqrt{x}\text{,}$

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\begin{align*} f'(x) \amp= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\\ \amp= \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \end{align*}

Then, the goal is to cancel the factor of $h$ in the denominator. It turns out that the correct next algebraic step is a technique called rationalizing the numerator. You may recall rationalizing the denominator, which involves multiplying numerator and denominator in order to eliminate a square root in the denominator. Here, we can use a similar technique with the goal of removing the square roots in the numerator. In this case, we multiply by the conjugate of the numerator, which is $\sqrt{x + h} + \sqrt{x}\text{,}$

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\begin{align*} \amp= \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}\\ \amp= \lim_{h \to 0} \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})}\\ \amp= \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}\\ \amp= \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} \amp\amp \text{cancelling the common factor of $h$}\\ \amp= \frac{1}{\sqrt{x} + \sqrt{x}} \amp\amp \text{plugging in $h = 0$}\\ \amp= \frac{1}{2\sqrt{x}} \end{align*}

Observe the relationship between $f(x)$ and its derivative: Desmos.

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The slope of $f(x) = \sqrt{x}$ starts out large and positive, and then as $x$ increases, the slope becomes smaller but still positive.
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This corresponds to the derivative $f'(x) = \frac{1}{2\sqrt{x}}$ being positive for all $x \gt 0\text{,}$ and getting closer to 0 as $x$ gets larger (or, $f'(x) \to 0$ as $x \to \infty$).
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