2.13: Definition of Derivative Examples

Last updated
Save as PDF

Page ID: 88637

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[1]/div/p[3]/span[1], line 1, column 1

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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[1]/div/p[3]/span[2], line 1, column 1

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

Now we continue to simplify and find the answer.

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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[1]/div/p[5]/span[1], line 1, column 1

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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[1]/div/p[5]/span[2], line 1, column 1

{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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[2]/div/p[2]/span[1], line 1, column 1

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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[2]/div/p[2]/span[2], line 1, column 1

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

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Calculus/Informal_Calculus_with_Applications_to_Biological_and_Environmental_Sciences_(Seacrest)/02:_Derivative_Introduction/2.13:_Definition_of_Derivative_Examples), /content/body/div[2]/div/p[2]/span[3], line 1, column 1

{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.