3.1 Defining the derivative (Page 3/10)

Calculus volume 1 Page 3 / 10

Finding the equation of a tangent line

Find the equation of the line tangent to the graph of $f (x) = 1 / x$ at $x = 2 .$

We can use [link] , but as we have seen, the results are the same if we use [link] .

\begin{matrix} m_{tan} & = lim_{x \to 2} \frac{f (x) - f (2)}{x - 2} & Apply the definition. \\ = lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} & Substitute f (x) = \frac{1}{x} and f (2) = \frac{1}{2} . \\ = lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} \cdot \frac{2 x}{2 x} & \begin{matrix} Multiply numerator and denominator by 2 x to \\ simplify fractions. \end{matrix} \\ = lim_{x \to 2} \frac{(2 - x)}{(x - 2) (2 x)} & Simplify. \\ = lim_{x \to 2} \frac{−1}{2 x} & Simplify using \frac{2 - x}{x - 2} = −1, for x \neq 2 . \\ = - \frac{1}{4} & Evaluate the limit. \end{matrix}

We now know that the slope of the tangent line is $- \frac{1}{4} .$ To find the equation of the tangent line, we also need a point on the line. We know that $f (2) = \frac{1}{2} .$ Since the tangent line passes through the point $(2, \frac{1}{2})$ we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation $y = - \frac{1}{4} x + 1 .$ The graphs of $f (x) = \frac{1}{x}$ and $y = - \frac{1}{4} x + 1$ are shown in [link] .

This figure consists of the graphs of f(x) = 1/x and y = -x/4 + 1. The part of the graph f(x) = 1/x in the first quadrant appears to touch the other function’s graph at x = 2. — The line is tangent to $f (x)$ at $x = 2 .$

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Find the slope of the line tangent to the graph of $f (x) = \sqrt{x}$ at $x = 4 .$

$\frac{1}{4}$

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The derivative of a function at a point

The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative . The process of finding a derivative is called differentiation .

Definition

Let $f (x)$ be a function defined in an open interval containing $a .$ The derivative of the function $f (x)$ at $a,$ denoted by $f^{'} (a),$ is defined by

f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}

provided this limit exists.

Alternatively, we may also define the derivative of $f (x)$ at $a$ as

f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} .

Estimating a derivative

For $f (x) = x^{2},$ use a table to estimate $f^{'} (3)$ using [link] .

Create a table using values of $x$ just below $3$ and just above $3 .$

$x$	$\frac{x^{2} - 9}{x - 3}$
$2.9$	$5.9$
$2.99$	$5.99$
$2.999$	$5.999$
$3.001$	$6.001$
$3.01$	$6.01$
$3.1$	$6.1$

After examining the table, we see that a good estimate is $f^{'} (3) = 6.$

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For $f (x) = x^{2},$ use a table to estimate $f^{'} (3)$ using [link] .

$6$

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Finding a derivative

For $f (x) = 3 x^{2} - 4 x + 1,$ find $f^{'} (2)$ by using [link] .

Substitute the given function and value directly into the equation.

\begin{matrix} f^{'} (x) & = lim_{x \to 2} \frac{f (x) - f (2)}{x - 2} & Apply the definition. \\ = lim_{x \to 2} \frac{(3 x^{2} - 4 x + 1) - 5}{x - 2} & Substitute f (x) = 3 x^{2} - 4 x + 1 and f (2) = 5 . \\ = lim_{x \to 2} \frac{(x - 2) (3 x + 2)}{x - 2} & Simplify and factor the numerator. \\ = lim_{x \to 2} (3 x + 2) & Cancel the common factor. \\ = 8 & Evaluate the limit. \end{matrix}

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Revisiting the derivative

For $f (x) = 3 x^{2} - 4 x + 1,$ find $f^{'} (2)$ by using [link] .

Using this equation, we can substitute two values of the function into the equation, and we should get the same value as in [link] .

\begin{matrix} f^{'} (2) & = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} & Apply the definition. \\ = lim_{h \to 0} \frac{(3 {(2 + h)}^{2} - 4 (2 + h) + 1) - 5}{h} & \begin{matrix} Substitute f (2) = 5 and \\ f (2 + h) = 3 {(2 + h)}^{2} - 4 (2 + h) + 1 . \end{matrix} \\ = lim_{h \to 0} \frac{3 h^{2} + 8 h}{h} & Simplify the numerator. \\ = lim_{h \to 0} \frac{h (3 h + 8)}{h} & Factor the numerator. \\ = lim_{h \to 0} (3 h + 8) & Cancel the common factor. \\ = 8 & Evaluate the limit. \end{matrix}

The results are the same whether we use [link] or [link] .

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For $f (x) = x^{2} + 3 x + 2,$ find $f^{'} (1) .$

$f^{'} (1) = 5$

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Velocities and rates of change

Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if $s (t)$ is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval $[a, t]$ if $t > a$ or $[t, a]$ if $t < a$ is given by the difference quotient

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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