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2.13: Definition of Derivative Examples

Last updated

Oct 19, 2021
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- 2.12: Algebra Tips and Tricks IV (Tips for dealing with fractions)
- 2.14: Project- Hard Definition of Derivative Problems

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In the last section, we saw the instantaneous rate of change, or derivative, of a function $f(x)$ at a point $x$ is given by

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

Definition of Derivative 1

Find the derivative of the function

$f(x) = 3x + 5$ using the definition of the derivative.

To use this in the formula $f'(x) = \frac{f(x+h) - f(x)}{h}$ , first we need to replace the $f(x+h)$ part of the formula. This is the same as $f(x)$ which is $3x+5$ , except we replace $x$ with that $(x+h)$ in parantheses. Like the following. The colors are only to highlight the substitution of $f(x+h)$ and $f(x)$ . We’ll drop the colors as soon as we need to combine expressions.

\[\begin{align*} f'(x) & = \lim_{h \to 0} \frac

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{h} \\ & = \lim_{h \to 0} \frac

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{h} \\ \end{align*}\]

Now we continue to simplify and find the answer.

\[\begin{align*} f'(x) & = \lim_{h \to 0} \frac

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{h} \\ & = \lim_{h \to 0} \frac

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{h} \\ & = \lim_{h \to 0} \frac{3x + 3h + 5 - 3x - 5}{h} \\ & = \lim_{h \to 0} \frac{3h}{h} \\ & = \lim_{h \to 0} 3 \\ & = \boxed{3} \end{align*}\]

Here, we have $f'(x) = 3$ . That makes sense if you think about it: $3x + 5$ is a line with slope $3$ !

Definition of Derivative 2

Find the derivative of

$f(x) = x^2$ using the definition.

\[\begin{align*} f'(x) & = \lim_{h \to 0} \frac

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{h} \\ & = \lim_{h \to 0} \frac

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{h} \\ & = \lim_{h \to 0} \frac

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{h} \\ & = \lim_{h \to 0} \frac{2xh + h^2}{h} \\ & = \lim_{h \to 0} \frac{h(2x + h)}{h} \\ & = \lim_{h \to 0} 2x + h \\ & = 2x + (0) \\ & = \boxed{2x} \end{align*}\]

So what does this mean? Well, this means we double $x$ to find the slope of the tangent line of $f(x) =x^2$ . So at $x = 3$ , the slope is $6$ , and at $x = 1.2$ , the slope is $2.4$ . ETC.

Definition of Derivative 1

Definition of Derivative 2

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