Module 08: The derivative and the tangent line

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: a stable limiting rate
C. Definitions and what they do not promise
D. Why a derivative implies continuity
E. An explicit routine for tangent problems
F. The transition you need most
Computing bridge: Two pieces of local information
G. Independent practice
H. Complete solutions: Module 08
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Module 05 difference quotients, Module 06 limits, and point-slope form.

After

Compute a derivative from its definition, separate function height from slope, build tangent lines, and distinguish differentiability from continuity.

B. The central idea: a stable limiting rate#

A secant line compares two points. A derivative asks whether these average slopes approach one finite number as the second point approaches the first from every allowed nearby direction. The derivative is not simply a tiny-interval quotient; it is the limit of such quotients when that limit exists.

At a fixed input $a$ , the derivative $f'(a)$ is a number. When the same process is carried out for varying inputs, $f'$ is itself a function. The prime symbol is an operator indicating differentiation, not an exponent. The notation $dy/dx$ is another way to denote the derivative of $y$ with respect to $x$ ; it is motivated by rates but should not initially be treated as an ordinary quotient of two nonzero numbers.

C. Definitions and what they do not promise#

A finite derivative at an interior point implies continuity there. The converse is false. Corners, cusps, and certain vertical tangencies can occur on continuous graphs without a finite derivative. An endpoint may have a one-sided derivative, which should be identified as such rather than silently described as a two-sided derivative. Definitions and the continuity implication: [O1D].

D. Why a derivative implies continuity#

For $h\ne0$ , rewrite the output change as

f(a+h)-f(a)=\left(\frac{f(a+h)-f(a)}h\right)h.

If the quotient tends to a finite $f'(a)$ , the product tends to $f'(a)\cdot0=0$ . Therefore $f(a+h)\to f(a)$ , giving continuity. The argument uses the existence of a finite quotient limit, not merely the fact that $h$ becomes small.

The derivative also produces a local prediction:

f(a+h)=f(a)+f'(a)h+\text{an error small compared with }|h|.

More precisely the error divided by $|h|$ tends to zero. This is the meaning of being locally well approximated by a line. The line need not equal the curve away from $a$ , and it need not meet the graph only once.

E. An explicit routine for tangent problems#

Find $f(a)$ to obtain the point. Find $f'(a)$ to obtain the slope. Place both into point-slope form before simplifying. Check that substituting $x=a$ gives $y=f(a)$ , and that the line’s coefficient of $x$ is the slope. If the derivative limit is not finite or the two sides disagree, state that no finite derivative exists rather than forcing a line of the form $y=mx+b$ .

F. The transition you need most#

For $f(x)=x^2-x-3$ , Module 05 derived the quotient $2x+h-1$ for nonzero $h$ . Now, and only now, take the limit:

f'(x)=\lim_{h\to0}(2x+h-1)=2x-1.

At $a=2$ , these are different evaluations:

f(2)=4-2-3=-1,\qquad f'(2)=4-1=3.

The point is $(2,-1)$ and the slope is $3$ . Thus

\begin{align*} y-(-1)&=3(x-2),\\ y+1&=3x-6,\\ \boxed{y}&\boxed{=3x-7}. \end{align*}

The $-1$ is a height at input $2$ ; the $-7$ is a height at input $0$ . The incorrect $y=3x-1$ would put the line through $(0,-1)$ , not through $(2,-1)$ .

A continuous counterexample. At zero, the quotient for $f(x)=|x|$ is $|h|/h$ . Its right-hand limit is $1$ and its left-hand limit is $-1$ . The derivative does not exist, even though Module 07 established continuity. The corner is a visual signal of the disagreeing limiting slopes, not a substitute for the calculation.

From a function graph to a derivative graph. A positive derivative means an upward-sloping tangent; a negative derivative means a downward-sloping tangent. On an interval where the derivative stays positive, the function increases; where it stays negative, it decreases (justified by the Mean Value Theorem in Module 13). The converse needs care: a strictly increasing function may have a horizontal tangent at an isolated point. A horizontal tangent corresponds to derivative zero. A function’s height and its derivative’s height are different: the graph can lie below the horizontal axis while having positive slope.

For

f(x)=x^3-3x

, the curve falls between

-1

and

1

and rises outside. Its horizontal tangencies occur at

x=\pm1

. The derivative

3x^2-3

has the corresponding negative and positive signs; do not confuse the curve’s zeros with its horizontal tangencies.

Computing bridge: Two pieces of local information#

A tangent approximation must preserve both a location and a rate. In engine terms, a moving object’s height at time $2$ is not its height at time $0$ . With $f(2)=-1$ and $f'(2)=3$ , the local prediction is $f(2+\Delta t)\approx-1+3\Delta t$ , which becomes $y=3x-7$ when $x=2+\Delta t$ . Substitution of $x=2$ checks the point independently of checking the slope. This same separation of baseline state and change appears in the integrator project J04.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M08.01FSolution

Use the definition to differentiate $f(x)=2x^2+3x-1$ . Show the entire difference quotient before simplification.

M08.02CSolution

For $f(x)=x^2-x-3$ , derive $f'(2)$ directly from $[f(2+h)-f(2)]/h$ , then find the tangent line at $2$ .

M08.03CSolution

Use a limit to find the derivative of $g(x)=1/x$ at a nonzero input $a$ .

M08.04TSolution

Let $f(4)=7$ and $f'(4)=-2$ . Find the tangent-line equation and its prediction for $f(4.03)$ . Is that prediction necessarily exact?

M08.05CSolution

Use the definition to decide whether $|x-2|$ is differentiable at $x=2$ . State its continuity status too.

M08.06TSolution

A graph is decreasing on $(-2,1)$ and has a horizontal tangent at $x=1$ . What can be concluded about its derivative where it exists? Does this determine the graph’s heights?

M08.07TSolution

Give a function with $f(0)=0$ and $f'(0)=2$ whose tangent line at zero intersects the graph at another point. Verify your example.

M08.08SSolution

Use the quotient at zero for $f(x)=x^{1/3}$ to explain why its graph has a vertical tangent but no finite derivative at zero.

Optional small hints

For M08.04, predict a change from the known point, not a new intercept. For M08.07, add to $2x$ a term vanishing at $0$ to second order and also at another input.

H. Complete solutions: Module 08#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M08.01Show solution

A general derivative from first principles

Substitute and subtract with brackets:

\frac{2(x+h)^2+3(x+h)-1-[2x^2+3x-1]}h.

Expansion gives numerator $2x^2+4xh+2h^2+3x+3h-1-2x^2-3x+1=4xh+2h^2+3h$ . Factor $h$ and cancel for $h\ne0$ to obtain $4x+2h+3$ . Taking the limit yields $\boxed{f'(x)=4x+3}$ . The cancellation and the limiting operation are distinct steps.

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M08.02Show solution

The point and slope reappear separately

Compute $f(2)=-1$ . Next $f(2+h)=(2+h)^2-(2+h)-3=-1+3h+h^2$ . Hence

\frac{f(2+h)-f(2)}h=\frac{[-1+3h+h^2]-[-1]}h=3+h.

The limit is $f'(2)=3$ . With point $(2,-1)$ the line is $y+1=3(x-2)$ , or $\boxed{y=3x-7}$ . At $x=2$ the result is $-1$ , verifying that the slope and point were combined correctly.

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M08.03Show solution

A reciprocal slope

For $a\ne0$ and sufficiently small nonzero $h$ with $a+h\ne0$ ,

\frac{1/(a+h)-1/a}h=-\frac1{a(a+h)}.

Its limit as $h\to0$ is $\boxed{-1/a^2}$ . A negative derivative is plausible: the reciprocal function decreases on each of its intervals $(-\infty,0)$ and $(0,\infty)$ .

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M08.04Show solution

Local prediction, not a new exact function

The tangent line is $y-7=-2(x-4)$ , hence $y=15-2x$ . For input change $0.03$ , its output change is $-2(0.03)=-0.06$ , giving $\boxed{f(4.03)\approx6.94}$ . It need not be exact because curvature can create an error. The given information determines the tangent line, not the whole function.

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M08.05Show solution

A translated corner

The value at $2$ is zero. Its difference quotient there is $|2+h-2|/h=|h|/h$ . The right limit is $1$ and the left limit $-1$ , so the derivative does not exist. Yet $|x-2|\to0=f(2)$ as $x\to2$ , so the function is continuous at $2$ .

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M08.06Show solution

Derivative signs do not encode absolute height

For a decreasing function, the derivative, where it exists, is nonpositive; strict decrease does not force it to be strictly negative at every individual point. The horizontal tangent gives $f'(1)=0$ if this tangent corresponds to a finite derivative. Adding a constant to the function shifts all heights without changing its slopes. Thus the stated behavior cannot determine function values.

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M08.07Show solution

Tangency does not mean only one intersection

Choose $f(x)=2x+x^2(x-1)$ . It satisfies $f(0)=0$ . Directly, $f(h)/h=2+h(h-1)\to2$ , so $f'(0)=2$ . The tangent is $y=2x$ . Since $f(1)=2$ , the graph meets this tangent again at $(1,2)$ . Indeed, equality with the line requires $x^2(x-1)=0$ , giving inputs $0$ and $1$ .

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M08.08Show solution

An infinite slope is not a finite derivative

At zero, the quotient is $h^{1/3}/h=1/h^{2/3}$ for $h\ne0$ . Because $h^{2/3}>0$ and tends to zero from both sides, this quotient tends to $+\infty$ . Thus there is no finite derivative at zero. The limiting lines become vertical, described geometrically by $x=0$ . The function is nevertheless continuous there because $h^{1/3}\to0$ .

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I. Retention and readiness#

Take mixed review R1 before rushing to derivative rules. Independently solve M05.02, M07.07, and M08.04. Explain in one sentence each what $f(a)$ and $f'(a)$ contribute to a tangent line. Advance when you can derive a fresh quadratic’s derivative and reject an incorrect tangent line by substitution.