5.8 Sketching Graphs of Functions and Their Derivatives

Every test in this unit links a feature of $f'$ or $f''$ to a feature of $f$. Sketching uses those links in both directions: derivative information produces a graph of $f$, and a graph of $f'$ describes $f$. Every sketch depends on correct derivatives, so this page starts with the differentiation rules.

Derivative rules that support the analysis

Accurate graph analysis depends on selecting the correct derivative rule. Familiar derivatives include

$$ \frac{d}{dx}(\sin x)=\cos x,\qquad \frac{d}{dx}(\cos x)=-\sin x,\qquad \frac{d}{dx}(e^x)=e^x. $$

For a product of differentiable functions,

$$ \frac{d}{dx}\bigl[f(x)g(x)\bigr] =f'(x)g(x)+f(x)g'(x). $$

For a quotient,

$$ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)

\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}, \qquad g(x)\ne 0. $$

The reciprocal identities make trigonometric derivatives manageable:

$$ \tan x=\frac{\sin x}{\cos x},\qquad \cot x=\frac{\cos x}{\sin x},\qquad \sec x=\frac{1}{\cos x},\qquad \csc x=\frac{1}{\sin x}. $$

They lead to

$$ (\tan x)'=\sec^2x,\qquad (\cot x)'=-\csc^2x, $$

$$ (\sec x)'=\sec x\tan x,\qquad (\csc x)'=-\csc x\cot x. $$

For instance, with $h(x)=x^2e^x$,

$$ h'(x)=2xe^x+x^2e^x=e^x(x^2+2x). $$

The first term differentiates the polynomial factor; the second differentiates the exponential factor.

Sketching a function and its derivatives

A reliable sketch combines several representations: a verbal description, a table of values, a graph of $f$, and graphs or sign information for $f'$ and $f''$. Technology can check a table or graph, but it does not replace explaining what the signs mean.

Information	Meaning for the graph of $f$
$f'(x)>0$	$f$ is increasing
$f'(x)<0$	$f$ is decreasing
$f'(x)=0$	Horizontal tangent or possible extremum
$f''(x)>0$	Slopes increase; graph bends upward
$f''(x)<0$	Slopes decrease; graph bends downward

The same table lets you read a graph of $f'$ directly. Where the graph of $f'$ lies above the $x$-axis, $f$ is increasing; where it lies below, $f$ is decreasing. Where $f'$ crosses the axis, $f$ has a relative extremum. Where the graph of $f'$ is rising, $f''>0$ and $f$ bends upward; where it is falling, $f$ bends downward. A turning point of $f'$ therefore marks an inflection point of $f$.

For $f(x)=x^3-3x$, the derivative graph $y=f'(x)=3x^2-3$ is an upward-opening parabola. Its zeros at $x=-1$ and $x=1$ are the horizontal tangencies of $f$; the sign pattern of $f'$ gives increasing, decreasing, and increasing intervals, while $f''(x)=6x$ identifies the local maximum and minimum through the Second Derivative Test.

This multirepresentational reasoning is Mathematical Practice 2: Connecting Representations and is assessed through FUN 2. Justification is Mathematical Practice 3: Justification, especially 3.D, when the response must explain why a point is a maximum or minimum rather than merely label it. Topic 5.7 is tagged FUN-4.A.7 for the Second Derivative Test and FUN-4.A.8 for distinguishing relative from absolute extrema.

Retrieval check. For $f(x)=x^3-3x$, the graph of $f'(x)=3x^2-3$ has its vertex at $x=0$. What does that vertex say about the graph of $f$? It marks an inflection point: $f'$ changes from decreasing to increasing there, so $f$ changes from bending downward to bending upward.