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Lesson 55 — Higher-Order Derivatives

Second derivative (concavity, acceleration), third derivative (jerk), nth-order formulas, inflection points, and a preview of Taylor series.

Used in: Cálculo I (Brasil) · Equiv. Math III japonês (cap. 4) · Equiv. Analysis LK alemão

f''(x) = \frac{d}{dx}\!\left[\frac{dy}{dx}\right] = \frac{d^2y}{dx^2}

The second derivative is the derivative of the derivative. Physically: if $f$ is position, then $f'$ is velocity and $f''$ is acceleration. Geometrically: $f'' > 0$ means the curve is concave up (a smile); $f'' < 0$ means the curve is concave down (a frown). Where $f''$ changes sign, there is an inflection point.

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous Definition

Higher-Order Derivatives

"If $y = f(x)$ , then the second derivative of $f$ is the derivative of $f'$ and is denoted $f''(x)$ or $d^2 y/dx^2$ . The process of calculating successive derivatives is called repeated differentiation." — OpenStax Calculus Vol. 1, §3.2

Equivalent Notations

Notation	Reading	Observation
$f''(x)$	"f double prime of x"	Newton; $n = 2$
$\dfrac{d^2y}{dx^2}$	"d squared y over d x squared"	Leibniz
$D^2 f$	"D squared f"	operational
$\ddot{y}$	"y double dot"	physics; independent variable is $t$
$f^{(n)}(x)$	"f nth of x"	general order
$\dfrac{d^n y}{dx^n}$	"d nth y"	general Leibniz

Table: Closed-form $n$ -th order formulas

$f(x)$	$f^{(n)}(x)$	Validity
$e^{ax}$	$a^n e^{ax}$	$a \in \mathbb{R}$ , $n \geq 0$
$\sin x$	$\sin\!\bigl(x + \tfrac{n\pi}{2}\bigr)$	$n \geq 0$
$\cos x$	$\cos\!\bigl(x + \tfrac{n\pi}{2}\bigr)$	$n \geq 0$
$x^k$	$\dfrac{k!}{(k-n)!} x^{k-n}$	$k \geq n$ ; zero if $k < n$
$\ln x$	$(-1)^{n-1}\dfrac{(n-1)!}{x^n}$	$x > 0$ , $n \geq 1$
$\dfrac{1}{x}$	$(-1)^n \dfrac{n!}{x^{n+1}}$	$x \neq 0$ , $n \geq 0$

Geometric Meaning — Concavity

"If $f''(x) > 0$ for all $x$ in $(a, b)$ , then $f$ is concave up on $(a, b)$ . If $f''(x) < 0$ for all $x$ in $(a, b)$ , then $f$ is concave down on $(a, b)$ ." — Active Calculus, §1.6

Concavity determined by the sign of f''. On the blue curve, f'' > 0 — the function "opens upwards". On the orange curve, f'' < 0 — the function "closes downwards".

Leibniz Rule for Product

$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)}\, g^{(n-k)}$

A perfect analogue of Newton's binomial theorem: replace power with the corresponding order derivative.

Taylor Polynomial of Degree $n$

T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k

what this means · Taylor polynomial of degree n around a. Each coefficient is determined by the k-th order derivative of f evaluated at a, divided by k factorial. It is the best polynomial approximation of f in the neighborhood of a.

Solved Examples

Example— 1· Second and third derivatives of a polynomial (application)

Problem. Let $f(x) = x^4 - 6x^3 + 5x + 1$ . Calculate $f'(x)$ , $f''(x)$ , and $f'''(x)$ .

Strategy. Differentiate successively, using the power rule $(x^n)' = nx^{n-1}$ at each step.

Solution.

Step 1 — First derivative:

$f'(x) = 4x^3 - 18x^2 + 5$

Step 2 — Second derivative (differentiate $f'$ ):

$f''(x) = 12x^2 - 36x$

Step 3 — Third derivative (differentiate $f''$ ):

$f'''(x) = 24x - 36$

Verification by degree count. $f$ has degree 4; $f'$ degree 3; $f''$ degree 2; $f'''$ degree 1; $f^{(4)} = 24$ ; $f^{(5)} = 0$ . Consistent.

Source. Active Calculus — Boelkins, §1.6 — CC-BY-NC-SA.

Example— 2· Acceleration from position (physical modeling)

Problem. The position of a particle (in meters) is given by $s(t) = 3t^3 - 12t^2 + 9t$ , $t \geq 0$ (seconds). Find velocity $v(t)$ , acceleration $a(t)$ , and jerk $j(t)$ . At what instant is the acceleration zero?

Strategy. $v = s'$ , $a = s'' = v'$ , $j = s''' = a'$ . Solve $a(t) = 0$ .

Solution.

$v(t) = s'(t) = 9t^2 - 24t + 9$

$a(t) = s''(t) = 18t - 24$

$j(t) = s'''(t) = 18 \text{ m/s}^3 \text{ (constant)}$

Zero acceleration: $18t - 24 = 0 \Rightarrow t = 4/3$ s.

For $t < 4/3$ : $a < 0$ (particle decelerates). For $t > 4/3$ : $a > 0$ (particle accelerates). The constant jerk indicates that the acceleration flow is uniform.

Source. APEX Calculus — Hartman, §2.2 — CC-BY-NC.

Example— 3· Pattern of the nth derivative of sin x (demonstrative)

Problem. Determine $(\sin x)^{(n)}$ for all $n \geq 0$ .

Strategy. Calculate the first derivatives and identify the cyclic pattern of period 4.

Solution.

$(\sin x)^{(0)} = \sin x$ $(\sin x)^{(1)} = \cos x = \sin\!\bigl(x + \tfrac{\pi}{2}\bigr)$ $(\sin x)^{(2)} = -\sin x = \sin\!\bigl(x + \pi\bigr)$ $(\sin x)^{(3)} = -\cos x = \sin\!\bigl(x + \tfrac{3\pi}{2}\bigr)$ $(\sin x)^{(4)} = \sin x = \sin\!\bigl(x + 2\pi\bigr)$

The cycle repeats. The general formula is:

$(\sin x)^{(n)} = \sin\!\left(x + \frac{n\pi}{2}\right)$

Verification: $(\sin x)^{(100)} = \sin(x + 50\pi) = \sin x$ , because $\sin$ has period $2\pi$ and $50\pi = 25 \cdot 2\pi$ .

Source. OpenStax Calculus Vol. 1, §3.2 — CC-BY-NC-SA.

Example— 4· Complete analysis of concavity and inflection points (intermediate)

Problem. For $f(x) = x^4 - 4x^3$ , determine the intervals of concavity and the inflection points.

Strategy. Calculate $f''$ , find zeros, create a sign table, identify sign changes.

Solution.

$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$ $f''(x) = 12x^2 - 24x = 12x(x - 2)$

Zeros of $f''$ : $x = 0$ and $x = 2$ .

Sign table:

Interval	Sign of $12x$	Sign of $(x-2)$	Sign of $f''$	Concavity
$(-\infty, 0)$	$-$	$-$	$+$	up
$(0, 2)$	$+$	$-$	$-$	down
$(2, +\infty)$	$+$	$+$	$+$	up

Inflection points: $x = 0$ and $x = 2$ (both with sign change of $f''$ ).

$f(0) = 0$ and $f(2) = 16 - 32 = -16$ .

Inflections: $(0, 0)$ and $(2, -16)$ .

Source. Active Calculus — Boelkins, §1.6 — CC-BY-NC-SA.

Example— 5· Taylor polynomial of degree 2 around a = 0 (preview)

Problem. Find the Taylor polynomial of degree 2 for $f(x) = e^x$ around $a = 0$ .

Strategy. Apply the formula $T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2$ .

Solution.

$(e^x)^{(n)} = e^x$ for all $n$ . Thus:

$f(0) = e^0 = 1, \quad f'(0) = 1, \quad f''(0) = 1$

$T_2(x) = 1 + x + \frac{x^2}{2}$

Numerical verification. For $x = 0{,}1$ : $e^{0{,}1} \approx 1{,}10517$ ; $T_2(0{,}1) = 1 + 0{,}1 + 0{,}005 = 1{,}105$ . Error $\approx 0{,}00017$ . Excellent for small $x$ .

Interpretation. The second derivative $f''(0) = 1 > 0$ guarantees that $e^x$ is concave up at $x = 0$ , which the paraboloid $T_2$ correctly captures.

Source. Active Calculus — Boelkins, §8.3 — CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 24Understanding 3Modeling 8Challenge 3Proof 2

Ex. 55.1Application
Let $f(x) = x^3 - 2x^2 + x - 5$ . Calculate $f'(x)$ and $f''(x)$ .
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First derivative: $f'(x) = 3x^2 - 4x + 1$ . Second derivative: $f''(x) = 6x - 4$ . The constant term disappears with each differentiation.
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1. Identify the function. $f(x) = x^3 - 2x^2 + x - 5$ . Goal: differentiate twice.
2. First derivative. Applying the power rule to each term: $f'(x) = 3x^2 - 4x + 1$ . The constant term $-5$ disappears.
3. Second derivative. Differentiating $f'$ : $f''(x) = 6x - 4$ . The constant term $1$ disappears.
4. Trick: each differentiation reduces the degree by 1 and multiplies by that degree. Quick check: degree of $f$ is 3, degree of $f''$ is 1. Correct.
Ex. 55.2Application
Let $f(x) = x^5 - 3x^2 + x + 2$ . Calculate $f''(x)$ .
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First derivative: $f'(x) = 5x^4 - 6x + 1$ . Second derivative: $f''(x) = 20x^3 - 6$ .
Ex. 55.3Application
Let $f(x) = \sin x$ . Calculate $f''(x)$ .
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First derivative: $f'(x) = \cos x$ . Second derivative: $f''(x) = -\sin x$ .
Ex. 55.4Application
Let $f(x) = \cos(2x)$ . Calculate $f''(x)$ .
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By the chain rule: $f'(x) = -2\sin(2x)$ . Second derivative: $f''(x) = -4\cos(2x)$ .
Ex. 55.5Application
Let $f(x) = \ln x$ . Calculate $f''(x)$ .
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$f(x) = \ln x$ has $f'(x) = x^{-1}$ . Second derivative: $f''(x) = -x^{-2} = -1/x^2$ .
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1. First derivative. $(\ln x)' = 1/x = x^{-1}$ .
2. Second derivative. Differentiating $x^{-1}$ by the power rule: $(-1) \cdot x^{-2} = -1/x^2$ .
3. Observation: $f''(x) < 0$ for all $x > 0$ — the logarithm is concave down throughout its domain.
Ex. 55.6ApplicationAnswer key
Let $f(x) = xe^x$ . Calculate $f''(x)$ .
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By the product rule: $f'(x) = e^x + xe^x = (1+x)e^x$ . Differentiating again: $f''(x) = e^x + (1+x)e^x = (2+x)e^x$ .
Ex. 55.7Application
Let $f(x) = x^2 \ln x$ . Calculate $f''(x)$ .
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By the product rule: $f'(x) = 2x\ln x + x^2 \cdot (1/x) = 2x\ln x + x$ . Differentiating again: $f''(x) = 2\ln x + 2x \cdot (1/x) + 1 = 2\ln x + 3$ .
Ex. 55.8Application
Let $f(x) = x^4 - 4x^3 + 1$ . Calculate $f'''(x)$ .
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$f'(x) = 4x^3 - 12x^2$ , $f''(x) = 12x^2 - 24x$ , $f'''(x) = 24x - 24$ .
Ex. 55.9ApplicationAnswer key
Let $f(x) = \dfrac{1}{1 + x^2}$ . Calculate $f''(0)$ .
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By the quotient rule: $f(x) = (1+x^2)^{-1}$ . $f'(x) = -2x(1+x^2)^{-2}$ . By the chain and product rules: $f''(x) = -2(1+x^2)^{-2} + 8x^2(1+x^2)^{-3}$ . At $x = 0$ : $f''(0) = -2 \cdot 1 + 0 = -2$ .
Ex. 55.10Application
Let $f(x) = \sqrt{x}$ . Calculate $f''(x)$ .
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$f(x) = x^{1/2}$ . $f'(x) = \frac{1}{2}x^{-1/2}$ . $f''(x) = -\frac{1}{4}x^{-3/2} = -\frac{1}{4x^{3/2}}$ . Note that $f'' < 0$ for all $x > 0$ — the square root is concave down.
Ex. 55.11ApplicationAnswer key
Let $f(x) = \cos(2x)$ . Calculate $f^{(4)}(x)$ .
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$f^{(1)} = -2\sin(2x)$ , $f^{(2)} = -4\cos(2x)$ , $f^{(3)} = 8\sin(2x)$ , $f^{(4)} = 16\cos(2x)$ . Each differentiation multiplies by 2 and advances the sine-cosine cycle.
Ex. 55.12Application
Let $f(x) = x^4$ . Calculate $f^{(5)}(x)$ .
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$f(x) = x^4$ is a polynomial of degree 4. The fifth derivative of any polynomial of degree 4 is zero, because $(x^4)^{(4)} = 4! = 24$ is constant and the derivative of a constant is zero.
Ex. 55.13Application
Let $f(x) = e^{2x}$ . Determine $f^{(n)}(x)$ for all $n \geq 1$ .
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Each differentiation of $e^{2x}$ by the chain rule multiplies by 2: $f^{(1)} = 2e^{2x}$ , $f^{(2)} = 4e^{2x}$ . In general $f^{(n)} = 2^n e^{2x}$ .
Ex. 55.14ApplicationAnswer key
Determine $(\sin x)^{(100)}$ .
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The sine function has a derivative cycle of period 4. $100 = 4 \times 25$ , remainder 0. Therefore $(\sin x)^{(100)} = (\sin x)^{(0)} = \sin x$ .
Ex. 55.15Application
Let $f(x) = \dfrac{1}{x}$ . Determine the general formula $f^{(n)}(x)$ .
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$f(x) = x^{-1}$ . $f^{(1)} = -x^{-2}$ , $f^{(2)} = 2x^{-3}$ , $f^{(3)} = -6x^{-4}$ . Pattern: $f^{(n)} = (-1)^n n! x^{-(n+1)}$ . Alternating sign comes from the product of consecutive negative factors.
Show step-by-step (with the why)
1. Calculate the first derivatives. $f = x^{-1}$ : $f' = -1 \cdot x^{-2}$ , $f'' = (-1)(-2)x^{-3} = 2x^{-3}$ , $f''' = (-1)(-2)(-3)x^{-4} = -6x^{-4}$ .
2. Identify the pattern. Coefficient of $f^{(n)}$ : $(-1)^n \cdot n!$ . Exponent: $-(n+1)$ .
3. Write the general formula. $f^{(n)}(x) = (-1)^n n! x^{-(n+1)}$ .
4. Curiosity: this formula is a special case of the derivative of $x^k$ : $(x^k)^{(n)} = k(k-1)\cdots(k-n+1)x^{k-n}$ , with $k = -1$ .
Ex. 55.16Application
For $f(x) = x^4 - 4x^3 + 1$ , determine the inflection points and intervals of concavity.
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$f''(x) = 12x^2 - 24x = 12x(x-2)$ . Zeros at $x = 0$ and $x = 2$ . $f'' > 0$ in $(-\infty, 0)$ , $f'' < 0$ in $(0, 2)$ , $f'' > 0$ in $(2, +\infty)$ . Both zeros are inflections (sign changes at each one).
Show step-by-step (with the why)
1. Calculate $f''$ . $f(x) = x^4 - 4x^3 + 1$ . $f' = 4x^3 - 12x^2$ . $f'' = 12x^2 - 24x = 12x(x-2)$ .
2. Zeros of $f''$ . $12x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$ .
3. Sign table. At $x = -1$ : $f'' = 36 > 0$ . At $x = 1$ : $f'' = -12 < 0$ . At $x = 3$ : $f'' = 36 > 0$ .
4. Conclusion. $f''$ changes sign at both points — inflections at $x = 0$ and $x = 2$ .
5. Mental shortcut: when $f''$ factors as a product of two distinct linear factors, each zero is automatically an inflection (sign alternates at each simple zero).
Ex. 55.17Application
For $f(x) = x^3 - 3x^2 + 2$ , determine the intervals of concavity and the inflection point.
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$f(x) = x^3 - 3x^2 + 2$ . $f'' = 6x - 6 = 6(x - 1)$ . For $x < 1$ : $f'' < 0$ (concave down). For $x > 1$ : $f'' > 0$ (concave up). Inflection at $x = 1$ , with $f(1) = 0$ .
Ex. 55.18ApplicationAnswer key
For $f(x) = e^{-x^2}$ , calculate $f''(0)$ .
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$f(x) = e^{-x^2}$ . By the chain rule: $f'(x) = -2x e^{-x^2}$ . By the product rule: $f''(x) = -2e^{-x^2} + (-2x)(-2x)e^{-x^2} = (-2 + 4x^2)e^{-x^2}$ . At $x = 0$ : $f''(0) = -2 \cdot 1 = -2$ .
Ex. 55.19Application
For $f(x) = x^5 - 5x^4$ , determine the inflection points.
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$f(x) = x^5 - 5x^4$ . $f' = 5x^4 - 20x^3$ . $f'' = 20x^3 - 60x^2 = 20x^2(x - 3)$ . Zeros: $x = 0$ (double) and $x = 3$ . At $x = -1$ , $f'' = 80 > 0$ ; at $x = 1$ , $f'' = -40 < 0$ ; at $x = 4$ , $f'' > 0$ . Sign change at $x = 0$ and $x = 3$ — both are inflections.
Ex. 55.20Understanding
If $f''(c) = 0$ , can we conclude that $c$ is an inflection point of $f$ ?
Select the correct option
Not necessarily — $f''(c) = 0$ is a necessary but not sufficient condition for inflectionYes, every zero of $f''$ is an inflection pointYes, if $f$ is a polynomialNo — inflection also requires $f'(c) = 0$
Select an option first
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Counterexample: $f(x) = x^4$ . $f''(x) = 12x^2$ and $f''(0) = 0$ . But $f'' \geq 0$ throughout $\mathbb{R}$ — it does not change sign. Sufficient condition for inflection: $f''$ changes sign when passing through $c$ .
Ex. 55.21Understanding
If $f'(c) = 0$ and $f''(c) > 0$ , what can be concluded about $c$ ?
Select the correct option
$c$ is a local minimum $c$ is a local maximum $c$ is an inflection pointThe test is inconclusive — cannot conclude
Select an option first
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If $f'(c) = 0$ and $f''(c) > 0$ , by the 2nd order Taylor expansion: $f(c+h) \approx f(c) + \frac{f''(c)}{2}h^2 > f(c)$ for small $h \neq 0$ . Thus $c$ is a local minimum. The positive curvature ensures that the function "opens upwards" around $c$ .
Ex. 55.22Application
Determine the concavity of $f(x) = e^x$ throughout its domain.
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$f(x) = e^x$ has $f''(x) = e^x > 0$ for all $x \in \mathbb{R}$ . Therefore $e^x$ is concave up throughout its domain. This implies that $e^x$ is a strictly convex function.
Ex. 55.23ApplicationAnswer key
Analyze the concavity of $f(x) = x^3$ and identify the inflection point.
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$f(x) = x^3$ . $f''(x) = 6x$ . For $x < 0$ : $f'' < 0$ (concave down). For $x > 0$ : $f'' > 0$ (concave up). Inflection at $x = 0$ — the point $(0, 0)$ is where the curve changes from a "frown" to a "smile".
Ex. 55.24Application
For $f(x) = x^4 - 6x^2$ , determine the intervals of concavity and the inflection points.
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$f(x) = x^4 - 6x^2$ . $f'' = 12x^2 - 12 = 12(x-1)(x+1)$ . Zeros: $x = \pm 1$ . $f'' > 0$ in $(-\infty, -1)$ and $(1, +\infty)$ ; $f'' < 0$ in $(-1, 1)$ . Inflections at $x = -1$ and $x = 1$ , with $f(-1) = f(1) = -5$ .
Ex. 55.25Understanding
Explain why $(\sin x)^{(4)} = \sin x$ and why $(e^x)^{(n)} = e^x$ for all $n \geq 0$ .
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For sine: differentiating four times cycles through $\sin \to \cos \to -\sin \to -\cos \to \sin$ , returning to the starting point — period 4. For the exponential: $(e^x)' = e^x$ — it is the only fixed point of the derivative operator (up to a multiplicative constant), therefore $(e^x)^{(n)} = e^x$ for all $n$ .
Ex. 55.26ApplicationAnswer key
Derive the formula for $(fg)''$ from the product rule, and identify the analogy with Newton's binomial theorem.
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By the product rule: $(fg)' = f'g + fg'$ . Differentiating again: $(fg)'' = f''g + f'g' + f'g' + fg'' = f''g + 2f'g' + fg''$ . The coefficient 2 in the mixed term is analogous to the binomial $(a+b)^2 = a^2 + 2ab + b^2$ .
Ex. 55.27Application
Let $f(x) = (1 + x)^{10}$ . Calculate $f^{(10)}(0)$ .
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$f(x) = (1+x)^{10}$ . The coefficient of the $x^{10}$ term in the binomial expansion is $\binom{10}{10} = 1$ . Therefore $f^{(10)}(x) = 10! \cdot 1 = 10!$ (constant). Evaluating at $x = 0$ : $f^{(10)}(0) = 10! = 3628800$ .
Ex. 55.28Modeling
Position of a particle: $s(t) = 4t^3 - t^4$ (meters, $t$ in seconds). Calculate $v(1)$ , $a(1)$ , and $j(1)$ , and interpret $j(1) = 0$ .
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$s(t) = 4t^3 - t^4$ . Velocity: $v(t) = 12t^2 - 4t^3$ ; $v(1) = 12 - 4 = 8$ m/s. Acceleration: $a(t) = 24t - 12t^2$ ; $a(1) = 24 - 12 = 12$ m/s². Jerk: $j(t) = 24 - 24t$ ; $j(1) = 0$ m/s³. The zero jerk at $t = 1$ indicates that the acceleration is at its maximum (inflection point of the acceleration curve).
Show step-by-step (with the why)
1. Velocity. $v(t) = s'(t) = 12t^2 - 4t^3$ . $v(1) = 12 - 4 = 8$ m/s.
2. Acceleration. $a(t) = v'(t) = 24t - 12t^2$ . $a(1) = 24 - 12 = 12$ m/s².
3. Jerk. $j(t) = a'(t) = 24 - 24t$ . $j(1) = 0$ m/s³.
4. Interpretation: jerk = 0 at $t = 1$ means that the acceleration is not changing at that instant — it has reached its extreme value.
Ex. 55.29Modeling
Pendulum: $\theta(t) = A\cos(\omega t)$ . Calculate $\ddot{\theta}$ and verify that $\ddot{\theta} + \omega^2\theta = 0$ .
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$\theta(t) = A\cos(\omega t)$ . $\dot\theta = -A\omega\sin(\omega t)$ . $\ddot\theta = -A\omega^2\cos(\omega t) = -\omega^2 \theta(t)$ . Therefore $\ddot\theta + \omega^2\theta = 0$ — equation of a simple pendulum (for small amplitudes). The second derivative of angular position is proportional to position, with a negative sign — principle of restoration.
Ex. 55.30Modeling
Production cost: $C(q) = q^3 - 6q^2 + 15q$ (R $thousand). Calculate$ C''(q)$ and interpret the inflection point as "minimum marginal cost".
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$C(q) = q^3 - 6q^2 + 15q$ . $C'(q) = 3q^2 - 12q + 15$ (marginal cost). $C''(q) = 6q - 12 = 6(q - 2)$ . Zero at $q = 2$ . For $q < 2$ : $C'' < 0$ (decreasing marginal cost). For $q > 2$ : $C'' > 0$ (increasing marginal cost). Inflection at $q = 2$ is the point of minimum marginal cost — maximum scale efficiency.
Show step-by-step (with the why)
1. Calculate $C''$ . $C'(q) = 3q^2 - 12q + 15$ . $C''(q) = 6q - 12$ .
2. Zero of $C''$ . $6q - 12 = 0 \Rightarrow q = 2$ .
3. Sign. $C'' < 0$ for $q < 2$ ; $C'' > 0$ for $q > 2$ . Sign change: inflection confirmed.
4. Economic interpretation: the marginal cost $C'$ reaches its minimum at $q = 2$ . Producing below 2 units: there is still a gain in scale. Above 2: marginal cost starts to rise.
Ex. 55.31ModelingAnswer key
Vehicle position: $s(t) = 10t^3 - 30t^2 + 5$ (meters). Calculate $v(t)$ , $a(t)$ , $j(t)$ and determine when acceleration is zero.
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$s(t) = 10t^3 - 30t^2 + 5$ . Velocity: $v(t) = 30t^2 - 60t$ . Acceleration: $a(t) = 60t - 60 = 60(t-1)$ . Jerk: $j(t) = 60$ m/s³ (constant). Acceleration zero at $t = 1$ s. For $t < 1$ : $a < 0$ ; for $t > 1$ : $a > 0$ . Constant jerk indicates linear variation of acceleration.
Ex. 55.32Modeling
Projectile height: $h(t) = -4{,}9t^2 + v_0 t + h_0$ . Calculate $h''(t)$ and identify its physical meaning.
Solve online OpenStax Calculus Vol. 1 · §3.2
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$h(t) = -4{,}9t^2 + v_0 t + h_0$ . $h'(t) = -9{,}8t + v_0$ (velocity). $h''(t) = -9{,}8$ m/s² — the gravitational acceleration, constant and negative (points downwards). The second derivative of position under gravity is always $-g$ , independent of $v_0$ and $h_0$ .
Ex. 55.33Modeling
In a mechanical system, the potential energy $U(\theta)$ has a critical point at $\theta_0$ . What does $U''(\theta_0) > 0$ versus $U''(\theta_0) < 0$ imply about the stability of the equilibrium?
Solve online APEX Calculus · §3.4
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In a system with potential energy $U(\theta)$ , equilibrium occurs where $U'(\theta_0) = 0$ . The second derivative test classifies: $U''(\theta_0) > 0$ — $U$ has a local minimum, stable equilibrium (system returns if perturbed). $U''(\theta_0) < 0$ — $U$ has a local maximum, unstable equilibrium (system escapes if perturbed). The second derivative quantifies the curvature of the potential, i.e., the effective stiffness of the system.
Ex. 55.34Modeling
Using the first three derivatives of $f(x) = e^x$ at $a = 0$ , write the Taylor polynomial $T_2(x)$ and estimate the error for $x = 0{,}1$ .
Solve online Active Calculus · §8.3
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$f(x) = e^x$ . $f(0) = 1$ , $f'(0) = 1$ , $f''(0) = 1$ . $T_2(x) = 1 + x + \frac{1}{2}x^2$ . For $x = 0{,}1$ : $T_2 = 1 + 0{,}1 + 0{,}005 = 1{,}105$ ; $e^{0{,}1} \approx 1{,}10517$ ; error $\approx 0{,}00017$ .
Ex. 55.35ModelingAnswer key
Write the Taylor polynomial of degree 2 for $f(x) = \cos x$ around $a = 0$ and verify for $x = 0{,}1$ .
Solve online Active Calculus · §8.3
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$f(x) = \cos x$ . $f(0) = 1$ , $f'(0) = -\sin(0) = 0$ , $f''(0) = -\cos(0) = -1$ . $T_2(x) = 1 + 0 \cdot x + \frac{-1}{2}x^2 = 1 - \frac{x^2}{2}$ . For $x = 0{,}1$ : $T_2 = 0{,}995$ ; $\cos(0{,}1) \approx 0{,}99500$ . Excellent approximation — error term is of order $x^4/24$ .
Ex. 55.36Challenge
Calculate $f^{(n)}(x)$ for $f(x) = \ln(1+x)$ and write the Taylor polynomial $T_n(x)$ around $a = 0$ .
Solve online Active Calculus · §8.3
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For $f(x) = \ln(1+x)$ : $f'(x) = (1+x)^{-1}$ , $f''(x) = -(1+x)^{-2}$ , $f'''(x) = 2(1+x)^{-3}$ . Pattern: $f^{(n)}(x) = (-1)^{n-1}(n-1)!(1+x)^{-n}$ . At $x = 0$ : $f^{(n)}(0) = (-1)^{n-1}(n-1)!$ . Therefore $\frac{f^{(n)}(0)}{n!} = \frac{(-1)^{n-1}}{n}$ , and $T_n(x) = \sum_{k=1}^n \frac{(-1)^{k+1}}{k}x^k$ .
Ex. 55.37Challenge
For $f(x) = x^x$ ( $x > 0$ ), calculate $f''(x)$ using logarithmic differentiation.
Solve online OpenStax Calculus Vol. 1 · §3.2
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Take logarithm: $\ln f = x\ln x$ . Differentiate: $f'/f = \ln x + 1$ , so $f' = x^x(1 + \ln x)$ . Differentiate again by the product rule: $f'' = x^x(1 + \ln x)^2 + x^x \cdot \frac{1}{x} = x^x(1 + \ln x)^2 + x^{x-1}$ .
Ex. 55.38Challenge
State Leibniz's formula $(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)} g^{(n-k)}$ and describe the structure of the induction argument that proves it.
Solve online APEX Calculus · §2.2
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See the referenced source for the detailed solution.
Ex. 55.39ProofAnswer key
Proof. Let $f$ be twice differentiable on $[0, 1]$ , with $f(0) = f(1) = 0$ and $f' \not\equiv 0$ . Does there exist $c \in (0, 1)$ with $f''(c) = 0$ ? Justify.
Solve online Active Calculus · §1.6
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Given $f(0) = f(1) = 0$ and $f$ differentiable, by Rolle's Theorem there exists $c_1 \in (0, 1)$ with $f'(c_1) = 0$ . Since $f' \not\equiv 0$ , there exists some point where $f' \neq 0$ . In particular, $f'$ has zeros (at least $c_1$ ). Applying Rolle's to $f'$ in an interval where $f'$ takes the same value at two points (or using MVT between a zero of $f'$ and a point where $f' \neq 0$ ), we obtain $c \in (0, 1)$ with $f''(c) = 0$ .
Ex. 55.40Proof
Proof. Prove that if $f$ is twice differentiable and $f''(x) \geq 0$ on $(a, b)$ , then $f$ is convex on $(a, b)$ .
Solve online Active Calculus · §1.6
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If $f'' \geq 0$ in $(a,b)$ , then $f'$ is non-decreasing. Given $x_1 < x < x_2$ in $(a,b)$ , write $x = \lambda x_1 + (1-\lambda)x_2$ with $\lambda \in (0,1)$ . By the MVT: $f(x_2) - f(x) \geq f'(x)(x_2 - x)$ and $f(x) - f(x_1) \leq f'(x)(x - x_1)$ (monotonicity of $f'$ ). Multiplying the first by $\lambda$ and the second by $1-\lambda$ and combining, we obtain $f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda)f(x_2)$ — definition of convexity.

Sources

Active Calculus 2.0 — Boelkins · 2024 · §1.6 (The Second Derivative), §8.3 (Taylor Polynomials). Primary source. CC-BY-NC-SA.
Calculus, Volume 1 — OpenStax · 2016 · §3.2 (The Derivative as a Function), §4.5 (Derivatives and the Shape of a Graph). CC-BY-NC-SA.
APEX Calculus — Hartman et al. · 2024 · v5 · §2.2 (Interpretations of the Derivative), §3.4 (Concavity and the Second Derivative). CC-BY-NC.