Exercises: Derivative function

Apply the definition of the derivative using the limit calculation $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ to find the derivative function of $f(x) = 3x^2 - 5x + 2$.

Calculate the derivative function of the following polynomial and exponential function, applying the basic rules of differentiation:

$f(x) = \frac{3}{4}x^4 - 2x^3 + \frac{5}{x^2} - 3e^x$

Obtain the derivative function of $f(x)$ applying the product rule. Simplify the result as much as possible.

$f(x) = (2x^3 - x + 4) \times \ln(x)$

Find the derivative function applying the quotient rule. Express the final result with the numerator simplified.

$f(x) = \frac{4x - 3}{x^2 + 5}$

Use the chain rule to determine the derivative function of the following composite trigonometric function:

$f(x) = \cos(5x^3 - 2x^2 + 1)$

Calculate the exact value of the derivative of the function $f(x) = \sqrt{3x^2 + 4}$ evaluated at the point with x-coordinate $x = 2$.

Determine the second derivative, $f''(x)$, of the function $f(x) = x^4 \times \ln(x)$.

Observe the graph of the function $f(x)$ representing the trajectory of a projectile as a function of time.

From the visual representation, determine the sign of the derivative function $f'(x)$ at the x-coordinates $x = 1$, $x = 2$, and $x = 3$. Justify your answer based on the geometric concept of the derivative.

Aoife is studying the relative extrema of the function $f(x) = (x-2)^3 + 1$. To do so, she computes its derivative function and obtains $f'(x) = 3(x-2)^2$. Next, she sets up the equation $f'(x) = 0$, finds $x = 2$ as the only solution, and concludes on her exam: "Since the derivative at $x = 2$ is zero, the function $f$ necessarily attains a relative minimum at the point $(2, 1)$".

1. a) Explain with reasons Aoife's conceptual error: does the condition $f'(a) = 0$ alone guarantee that a relative extremum exists at $x = a$?
2. b) Study the sign of the derivative function $f'(x)$ on the intervals $(-\infty, 2)$ and $(2, +\infty)$. From that analysis, describe what the curve actually does to the left and to the right of $x = 2$, and deduce what type of point the function has at $(2, 1)$.

Consider the polynomial function $f(x) = 4 - x^2$.

1. a) Before computing any derivative, make a reasoned qualitative estimate: what sign do you expect the slope of the tangent line to the curve to have at $x = 1$ and at $x = -2$? At which of the two points do you anticipate the steepness to be greater in absolute value? Justify your answers based on the increasing/decreasing behavior and the symmetric shape of the parabola.
2. b) Obtain the general derivative function $f'(x)$ by rigorously applying the definition of derivative using the limit of the difference quotient: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
3. c) Evaluate $f'(1)$ and $f'(-2)$ by substituting into the derivative function you obtained, and check whether the numerical results confirm the estimates you formulated in part a).

The equation of motion of an autonomous test vehicle is given by the function $s(t) = 2t^3 - 15t^2 + 24t + 10$, where $s$ is the position in metres and $t$ is the time in seconds ($t \ge 0$).

Find the velocity function $v(t)$ and the acceleration function $a(t)$. Determine at what times the vehicle comes to a momentary standstill.

The total production cost, in euros, of $x$ units of a high-precision electronic component is modelled by the function $C(x) = 0.02x^3 - 0.6x^2 + 40x + 500$.

Calculate the marginal cost function $C'(x)$ and evaluate its value when exactly $x = 15$ units are produced. Interpret the result obtained in the context of the problem.

In a chemical kinetics experiment, the concentration of a reactant as a function of time $t$ (in minutes) fits the expression $C(t) = \frac{8t}{t^2 + 4}$ in moles per litre.

Find the derivative function $C'(t)$ that represents the reaction rate. At what time $t > 0$ is the reaction rate zero?

Determine the values of the real parameters $a$ and $b$ such that the following piecewise-defined function is differentiable at $x = 2$:

$f(x) = \begin{cases} ax^2 + 3x & \text{if } x \le 2 \\ bx^3 - 4 & \text{if } x > 2 \end{cases}$

Apply the properties of logarithms to simplify the expression for $f(x)$ before differentiating, and then calculate its simplified derivative function:

$f(x) = \ln\left( \sqrt{\frac{1 + \sin(x)}{1 - \sin(x)}} \right)$

Find the x-coordinates of the points on the curve $f(x) = e^{3x}(x^2 - 4x + 2)$ at which the tangent line is horizontal.

The cumulative number of views of a viral video $t$ hours after its publication follows an exponential growth law given by $R(t) = 1500 \times 3^{0.4t}$.

Obtain the function describing the instantaneous growth rate of views, $R'(t)$, and calculate this rate after $5$ hours.

The following graph represents the derivative function $f'(x)$ of a continuous function $f(x)$ across all of $\mathbb{R}$.

Observing exclusively the behaviour of $f'(x)$, determine the intervals where the original function $f(x)$ is increasing and where it is decreasing. At which x-coordinates will $f(x)$ have relative extrema?

Determine the coefficients $a$, $b$, and $c$ of a quadratic function $f(x) = ax^2 + bx + c$ knowing that it simultaneously satisfies the following three conditions:

1. Its second derivative function is constant and equal to $-4$ throughout its domain: $f''(x) = -4$.

2. Its derivative function vanishes at the abscissa $x = 1.5$, being strictly positive for $x < 1.5$ and strictly negative for $x > 1.5$.

3. The graph of $f$ passes through the point $(1.5; 6)$.

Once you have calculated the expressions for $f(x)$, $f'(x)$, and $f''(x)$:

1. a) Justify with reasons, based on the sign change of $f'(x)$, whether the point $(1.5; 6)$ is a relative maximum or a relative minimum.
2. b) Explain what the value and sign of the second derivative $f''(x)$ mean regarding how fast the slope of the function changes around that extremum.

During a heatwave day recorded by Met Éireann at a weather station in Wexford, in the south-east of Ireland, the outside temperature $T$ (in degrees Celsius) between 08:00 ($t = 0$) and 20:00 ($t = 12$) is modeled by the function:

where $t$ is the time measured in hours elapsed since 08:00.

1. a) Obtain the derivative function $T'(t)$ and interpret what physical quantity it represents in this context, stating its corresponding units of measurement.
2. b) Calculate the instantaneous rate of change of the temperature at 10:00 ($t = 2$) and at 18:00 ($t = 10$). Interpret the physical meaning of the sign and the value obtained in each case.
3. c) Solve the equation $T'(t) = 0$ to find the exact time at which the temperature reaches its daily maximum value, and calculate that maximum temperature.
4. d) Determine the second derivative function $T''(t)$. What does this second derivative measure regarding the progression of the heat throughout the day, and what does its negative sign throughout the interval tell us?

Find the general expression for the nth derivative, $f^n(x)$, of the function $f(x) = \ln(3x - 2)$. Hint: calculate the first four successive derivatives and deduce the pattern.

Consider the function defined using a modulus $f(x) = |x^2 - 4|$.

Express $f(x)$ as a piecewise function and calculate its derivative function $f'(x)$. Prove analytically why the function is not differentiable at the corner points indicated on the graph.

Show that the function $y = e^{-2x} \times \sin(3x)$ is a particular solution to the second-order linear differential equation:

$y'' + 4y' + 13y = 0$

Given the rational function $f(x) = \frac{ax^2 + b}{cx + 1}$, determine the values of the real parameters $a, b, c$ knowing that its graph passes through the point $(0, 3)$, has a vertical asymptote at $x = -2$, and has a horizontal tangent line at the point with x-coordinate $x = 1$.

In the design of a suspension bridge, the profile of the main cable takes the form of a catenary whose equation in a coordinate system is $f(x) = \frac{e^{x/10} + e^{-x/10}}{2}$ (with $x$ and $y$ in metres).

Calculate the angle of inclination $\alpha$ that the tangent line to the cable makes with the horizontal at the anchorage point located at $x = 10$ metres.

Consider the function $f(x) = \begin{cases} e^{ax} - 1 & \text{if } x \le 0 \\ b \sin(x) + c x^2 & \text{if } x > 0 \end{cases}$.

Determine the mathematical relationship that must exist between the parameters $a, b, c$ for the second derivative to exist at the origin, that is, for $f''(0)$ to be a well-defined real value.

The following panel shows two curves, A (blue) and B (green). It is known that one of them corresponds to a polynomial function $f(x)$ and the other to its derivative function $f'(x)$.

Give reasons to identify which curve represents $f(x)$ and which represents $f'(x)$ based on the geometric relationship between the intercepts, the relative extrema, and the intervals of increase/decrease. Next, deduce the sign of the second derivative $f''(x)$ at the x-coordinate $x = 0$.

The goal is to determine the derivative function and the intervals of increase and decrease of the rational function $f(x) = \frac{2x+1}{x-3}$, whose domain is $\mathbb{R} \setminus \{3\}$. To calculate $f'(x)$ using the limit definition, two students propose two different working methods:

• Strategy A (Direct difference quotient): Calculate $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{2(x+h)+1}{x+h-3} - \frac{2x+1}{x-3}}{h}$ by finding a common denominator directly in the numerator.

• Strategy B (Prior decomposition): Perform polynomial division to write $f(x) = 2 + \frac{7}{x-3}$ and apply the limit definition to this new expression: $f'(x) = \lim_{h \to 0} \frac{\left(2 + \frac{7}{x+h-3}\right) - \left(2 + \frac{7}{x-3}\right)}{h}$.

1. a) Compare both strategies and explain which one is more algebraically efficient and less prone to simplification errors in the limit calculation.
2. b) Fully carry out the calculation of $f'(x)$ following the strategy you selected as more advantageous.
3. c) From the expression for $f'(x)$, specify the domain of differentiability of $f$, analyze the sign of $f'(x)$ on the intervals $(-\infty, 3)$ and $(3, +\infty)$, and explain why the function has no relative extrema despite being unbounded.

To study the differentiability and relative extrema of the function $f(x) = |x^2 - 4|$, Cian writes on his exam:

"I write the function as a piecewise function: $f(x) = x^2 - 4$ if $x \le -2$ or $x \ge 2$, and $f(x) = 4 - x^2$ if $-2 < x < 2$. I differentiate each piece directly: $f'(x) = 2x$ on the outside and $f'(x) = -2x$ on the inside. Since evaluating in the first formula gives $f'(2) = 4$ and $f'(-2) = -4$, the function is differentiable on all of $\mathbb{R}$. Then I solve $f'(x) = 0$: the only valid solution in its interval is $-2x = 0 \implies x = 0$. In conclusion, the only relative extremum of $f$ is the relative maximum at $x = 0$".

1. a) Identify and explain in detail the two conceptual errors made by Cian: one regarding how the existence of the derivative function at the boundary points ($x = -2$ and $x = 2$) is determined, and another regarding the complete search for relative extrema.
2. b) Apply the definition of one-sided derivatives (using one-sided limits of the difference quotient) at $x = 2$ to prove whether $f'(2)$ exists or not. What geometric feature does the curve exhibit at $x = 2$ and at $x = -2$?
3. c) Correct the exercise by determining the exact domain of differentiability of $f(x)$ and identifying ALL of its relative extrema (classifying them as maxima or minima). Explain how it is possible for relative extrema to exist at points where the function is not differentiable.