<caption>The derivative <m>f'(x)</m> is positive everywhere because the function <m>f(x)</m> is increasing.</caption>
<caption>The derivative <m>f'(x)\lt 0</m> where the function <m>f(x)</m> is decreasing and <m>f'(x)\gt 0</m> where <m>f(x)</m> is increasing...</caption>
<description>The function f(x) = the square root of x is graphed as is its derivative f'(x) = 1/(2 times the square root of x).</description>
<description>Two functions are graphed here: f(x) and f'(x). The function f(x) is the same as the above graph, that is, roughly sinusoidal, starting at (-4, 3), decreasing to a local minimum at (-2, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2). The function f'(x) is an downward-facing parabola with vertex near (0.5, 1.75), y-intercept (0, 1.5), and x-intercepts (-1.9, 0) and (3, 0).</description>
<description>The function f(x) = x squared - 2x is graphed as is its derivative f'(x) = 2x - 2.</description>
<description>The function f(x) is roughly sinusoidal, starting at (-4, 3), decreasing to a local minimum at (-2, 2), then increasing to a local maximum at (3, 6), and getting cut off at (7, 2).</description>
<example></example>
<exercise></exercise>
<figure></figure>
<figure></figure>
<image source="solution-image.jpg"></image>
<image source="fx-sqrt-x.jpg"></image>
<image source="exercise-image.jpg"></image>
<image source="fx-x2-2x.jpg"></image>
<p>We have already discussed how to graph a function...</p>
<p>In Example 3.12 we found that for <m>f(x)=x^2-2x,f'(x)=2x-2</m>...</p>
<p>Sketch the graph of <m>f(x)=x^2-4</m>...</p>
<p>In Example 3.11 we found that for <m>f(x)=\sqrt{x},f'(x)=1/2\sqrt{x}</m>...</p>
<p>The solution is shown in the following graph. Observe that...</p>
<p>Use the following graph of <m>f(x)</m> to sketch a graph of <m>f'(x)</m>.</p>
<solution></solution>
<statement></statement>
<statement></statement>
<subsection></subsection>
<title>Graphing a Derivative</title>
<title>Sketching a Derivative Using a Function</title>