Derivatives

(cod: P-49-43-6) We say that a function \(f\) is differentiable or derivable at a number \(x_0\) in its domain if there exists \begin{equation}\label{eq1} \lim\limits_{h \to 0} \dfrac{f(x_0 + h) - f(x_0)}{h}\end{equation} and, in this case, the derivative of \(f\) at \(x_0\), denoted by \(f'(x_0)\), is given by the value of this limit, that is,

\[f'(x_0) = \lim\limits_{h \to 0} \dfrac{f(x_0 + h) - f(x_0)}{h}.\] For a fixed value of \(h\), the ratio \[\dfrac{f(x_0 + h) - f(x_0)}{h}, \] appearing in the limit (\ref{eq1}) can be interpreted as the slope (inclination, or angular coefficient) of the secant line passing through the points \(P = (x_0+h,f(x_0+h))\) and \(P_0 = (x_0,f(x_0))\). If \(f\) is differentiable at \(x_0\), the line passing through \(P_0 = (x_0,f(x_0))\) with slope \(f'(x_0)\) is called the tangent line to the graph \(y = f(x)\) at \(P_0\). In the animation below, the tangent line at \(P_0\) appears in lilac, in a case where \(x_0 = 1\).

We observe that the limit in (\ref{eq1}) is equivalent to the limit \begin{equation}\label{eq3} \lim\limits_{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0},\end{equation} that is, if one exists, then the other also exists and has the same value.

Given a function \(f: U \subset \mathbb{R} \to \mathbb{R}\), we can consider the derivative of \(f\) as the function \[f'(x) = \lim\limits_{h \to 0} \dfrac{f(x+h) - f(x)}{h},\] defined on the set \[ \{x \in U: \ f \text{ is differentiable at } x\}.\] Given a function \(y = f(x)\), we can use another way to denote its derivative, known as Leibniz notation (note that in this notation, we use \(\Delta x\) instead of \(h\)): \[\dfrac{dy}{dx} = \lim \limits_{\Delta x \to 0} \dfrac{\overbrace{f(x+\Delta x) - f(x)}^{\Delta y}}{\Delta x}.\] In fact, there are several ways to denote the derivative of a function \(y = f(x)\). We present some below: \[f'(x) = \dfrac{dy}{dx} = \dfrac{d}{dx} \left(f(x)\right) = Df(x).\] Leibniz notation emphasizes the derivative's nature as a rate of change. If \(y = f(x)\), note that \[\dfrac{\Delta y}{\Delta x} = \dfrac{f(x+\Delta x) - f(x)}{\Delta x}\] represents an average rate of change of \(y\) with respect to \(x\). The derivative arises when we let \(\Delta x\) approach zero. Let us consider a more specific example. Suppose an object performs a rectilinear motion, so that its position can be described by a function \(s = s(t)\) and represented graphically as a point on an axis. Note that \[\dfrac{\Delta s}{\Delta t} = \dfrac{s(t+\Delta t) - s(t)}{\Delta t}\] represents the average velocity over the time interval between the instants \(t\) and \(t + \Delta t\). In this case, the derivative of the position function represents the instantaneous velocity of the object at the instant \(t\): \[\small{v(t) = \dfrac{ds}{dt} = \lim\limits_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t} = \lim\limits_{\Delta t \to 0}\dfrac{s(t+\Delta t) - s(t)}{\Delta t}.}\]

We have seen that, by differentiating a function, we obtain a new function, which is its derivative. We can then differentiate this new function, obtaining the second derivative or second derivative or derivative of second order. The second derivative of a function \(y = f(x)\) is denoted by \[f''(x) \ \ \text{ or } \ \ \dfrac{d^2 y}{dx^2} \ \ \text{ or } \ \ f^{(2)}(x).\] Proceeding inductively, if \(k\) is a natural number, we can consider the \(k\)-th derivative or derivative of order \(k\) of \(y = f(x)\), which we denote by \[f^{(k)}(x) \ \ \text{ or } \ \ \dfrac{d^k y }{dx^k}.\] If \(s = s(t)\) describes the position of an object in rectilinear motion, then the second derivative \(s''(t)\) represents the instantaneous acceleration of the object at the instant \(t\).

Consider the following questions:

(1) If \(f\) is a differentiable function at a number \(x_0\) in its domain, then is \(f\) necessarily continuous at \(x_0\)? Conversely, if a function \(f\) is continuous at a number \(x_0\) in its domain, then is \(f\) necessarily differentiable at \(x_0\)?

(2) If \(n\) is a natural number, determine the tangent line to the graph of the function \(f(x) = x^n\) at the point \(P_0 = (1,1)\).

(3) A projectile is launched vertically upwards. Its height, in feet, after \(t\) seconds, is given by \(s = s(t) = 144t - 16t^2, \ t \in [0,9]\). Determine the velocity \(v(t)\) of the projectile (in feet per second) and the acceleration \(a(t)\) of the projectile (in feet per second squared).

Select the correct alternative:

(1) If a function is differentiable at \(x_0\), then it is necessarily continuous at \(x_0\). Conversely, if a function is continuous at a number \(x_0\), then \(f\) is differentiable at \(x_0\).

(2) \(y = n(x-1) - 1.\)

(3) \begin{eqnarray*} v(t) &=& \dfrac{ds}{dt} &=& 144 - 32t, \\ & & \\ a(t) &=& \dfrac{d^2 s}{dt^2} &=& -32t. \end{eqnarray*}

(1) If a function is continuous at \(x_0\), then it is necessarily differentiable at \(x_0\). However, there exist functions that are differentiable at a number \(x_0\) but not continuous at \(x_0\).

(2) \(y = n(x-1) + 1.\)

(3) \begin{eqnarray*} v(t) &=& \dfrac{ds}{dt} &=& 144 - 16t, \\ & & \\ a(t) &=& \dfrac{d^2 s}{dt^2} &=& -16. \end{eqnarray*}

(1) If a function is differentiable at \(x_0\), then it is necessarily continuous at \(x_0\). However, there exist functions that are continuous at a number \(x_0\) but not differentiable at \(x_0\).

(2) \(y = n(x-1) + 1.\)

(3) \begin{eqnarray*} v(t) &=& \dfrac{ds}{dt} &=& 144 - 16t, \\ & & \\ a(t) &=& \dfrac{d^2 s}{dt^2} &=& -16. \end{eqnarray*}

(1) If a function is differentiable at \(x_0\), then it is necessarily continuous at \(x_0\). However, there exist functions that are continuous at a number \(x_0\) but not differentiable at \(x_0\).

(2) \(y = n(x-1) - 1.\)

(3) \begin{eqnarray*} v(t) &=& \dfrac{ds}{dt} &=& 144 - 32t, \\ & & \\ a(t) &=& \dfrac{d^2 s}{dt^2} &=& -32t. \end{eqnarray*}

(1) If a function is differentiable at \(x_0\), then it is necessarily continuous at \(x_0\). However, there exist functions that are continuous at a number \(x_0\) but not differentiable at \(x_0\).

(2) \(y = n(x-1) + 1.\)

(3) \begin{eqnarray*} v(t) &=& \dfrac{ds}{dt} &=& 144 - 32t, \\ & & \\ a(t) &=& \dfrac{d^2 s}{dt^2} &=& -32. \end{eqnarray*}

(cod: P-49-28-5) Below are statements involving the derivative of various functions:

(1) If \(c \in \mathbb{R}\), \(\dfrac{d}{dx}\left(c\right) = 0.\)

(2) If \(c \in \mathbb{R}\), \(\dfrac{d}{dx}\left(cx\right) = cx\)

(3) If \(n \in \mathbb{N}\), \(\dfrac{d}{dx}\left(x^n \right) = nx^{n-1}\)

(4) \(\dfrac{d}{dx}\left(x^{1/2}\right) = \frac{1}{2} x^{1/2 - 1} = \dfrac{1}{2\sqrt{x}}\)

(5) If \(n \in \mathbb{N}\), \(\dfrac{d}{dx}\left(x^{-n} \right) = nx^{-n-1}\)

(6) \(\dfrac{d}{dx}\left(\sin(x) \right)= \cos(x) \)

(7) \(\dfrac{d}{dx}\left(\cos(x)\right)= \sin(x)\)

(8) \(\dfrac{d}{dx}\left(\tan(x)\right) = \sec^2(x)\)

(9) \(\dfrac{d}{dx}\left(\sec(x)\right) = \tan^2(x)\)

(10) \(\dfrac{d}{dx}\left(b^x \right) = b^x \ln(b)\)

(11) If \(0 < b \ne 1\), \(\dfrac{d}{dx}\left(\log_b(x)\right) = \dfrac{1}{x}\)

(12) \(\dfrac{d}{dx}\left(e^x\right) = e^x\)

(13) \(\dfrac{d}{dx}\left(\ln(x)\right) = \dfrac{1}{x}\)

(14) \(\dfrac{d}{dx}\left(3^x\right) = x 3^{x-1}\)

We can state that:

Among these, the only statements that are false are numbers (2), (10), and (11).

Among these, the only statements that are false are numbers (4), (8), (10), and (11).

Among these, the only statements that are false are numbers (2), (4), (7), and (14).

Among these, the only statements that are false are numbers (2), (5), (7), (9), (11), and (14).

Among these, the only statements that are false are numbers (2), (5), (8), (9), and (14).

(cod: P-52-29-11) Consider the following questions:

(1) \(\dfrac{d}{dx}\left(\dfrac{2x^2}{x^2 - 9}\right)\) is equal to:

(2) \(\dfrac{d}{dx}\left(e^{-2x}\cos(3x)\right)\) is equal to:

(3) Let \(F\), \(G\), and \(H\) be differentiable functions such that: \begin{eqnarray*} F(0) &=& 2, \ \ \ \ \ \ \ F'(0) &=& 3, \\ G(0) &=& 1, \ \ \ \ \ \ \ G'(0) &=& -1, \\ G(1) &=& -1, \ \ G'(1) &=& 1, \\ H(0) &=& 1, \ \ \ \ \ \ \ H'(0) &=& 2. \end{eqnarray*} If \(f(x) = F(x)G(H(x))\), determine \(f'(0)\).

(4) Let \(f : I \to J\) be a differentiable bijection between open intervals \(I\) and \(J\) such that \[f(1) = 0 \ \ \ \text{and} \ \ \ f'(1) = \dfrac{1}{2}.\] Denote by \(f^{-1}:J \to I\) the inverse of \(f\) and consider the function \(g(t) = (t^2 + t + 3)f^{-1}(t)\). Determine the value of \(g'(0)\).

Select the alternative that correctly answers the four questions above.

\begin{eqnarray*} &(1)& - \dfrac{36x}{(x^2 - 9)^2} \\ & & \\ &(2)& e^{-2x}\left(\cos(3x) + \sin(3x)\right) \\ & & \\ &(3)& f'(0) = -1 \\ & & \\ &(4)& g'(0) = 7 \end{eqnarray*}

\begin{eqnarray*} &(1)& - \dfrac{36x}{(x^2 - 9)^2} \\ & & \\ &(2)& e^{-2x}\left(\cos(3x) + \sin(3x)\right) \\ & & \\ &(3)& f'(0) = -1 \\ & & \\ &(4)& g'(0) = 6 \end{eqnarray*}

\begin{eqnarray*} &(1)& 2 \\ & & \\ &(2)& - e^{-2x}\left(2\cos(3x) + 3 \, \sin(3x)\right) \\ & & \\ &(3)& f'(0) = -1 \\ & & \\ &(4)& g'(0) = 6 \end{eqnarray*}

\begin{eqnarray*} &(1)& 2 \\ & & \\ &(2)& e^{-2x}\left(\cos(3x) + \sin(3x)\right) \\ & & \\ &(3)& f'(0) = 1 \\ & & \\ &(4)& g'(0) = 7 \end{eqnarray*}

\begin{eqnarray*} &(1)& - \dfrac{36x}{(x^2 - 9)^2} \\ & & \\ &(2)& - e^{-2x}\left(2\cos(3x) + 3 \, \sin(3x)\right) \\ & & \\ &(3)& f'(0) = 1 \\ & & \\ &(4)& g'(0) = 7 \end{eqnarray*}

(cod: P-52-32-5) Consider the following statements:

(1) If \(f(x) = x^r\), for \(x > 0\), where \(r\) is any real number, then \(f'(x) = r x^{r-1}\).

(2) \(\dfrac{d}{dx}\ln(|x|) = \dfrac{1}{x}.\)

(3) \(\dfrac{d}{dx} \sinh\, x = \cosh{x}.\)

(4) \(\dfrac{d}{dx} \cosh{x} = - \sinh\, {x}.\)

(5) \(\dfrac{d}{dx} \tanh{x} = \operatorname{sech}^2\, {x}.\)

(6) \(\dfrac{d}{dx} \operatorname{sech}\, {x} = \operatorname{sech}\, {x} \tanh{x}.\)

Regarding these statements, we have that:

The only false statements are (4) and (6).

The only false statements are (2) and (3).

The only false statements are (1) and (5).

The only false statement is (2).

The only false statements are (2) and (4).

(cod: P-53-35-12) Consider the curve \(\mathcal{C}\) given by the equation \[x^2 + 2xy + 3y^2 = 8.\] We take a circle \(C_r\) with radius \(r>0\) centered at the origin such that it contains the curve \(\mathcal{C}\) within its interior. Then, we gradually decrease \(r\) until the circle touches the curve \(\mathcal{C}\) at two points \(P_1\) and \(P_2\), as shown in the animation below.

At points \(P_1\) and \(P_2\), the circle and the curve \(\mathcal{C}\) are tangent, meaning that at each of these points, the curve \(\mathcal{C}\) and the circle touching it share the same tangent line. Observe that \(P_1\) and \(P_2\) are the points on the curve \(\mathcal{C}\) farthest from the origin. Let \(D_M\) denote the distance of these points from the origin.

Now, we take a circle with radius \(r>0\) centered at the origin and contained within the region bounded by the curve \(\mathcal{C}\). Then, we gradually increase \(r\) until the circle touches the curve \(\mathcal{C}\) at points \(P_3\) and \(P_4\), as shown in the animation below.

At points \(P_3\) and \(P_4\), the circle and the curve are tangent. Note that \(P_3\) and \(P_4\) are the points on the curve \(\mathcal{C}\) closest to the origin. Let \(D_m\) denote the distance of these points from the origin.

Using implicit differentiation and the observations above, we can verify that the values of \(D_M\) and \(D_m\) are, respectively:

\(\sqrt{8 \sqrt{2} + 2}\) and \(\sqrt{8 \sqrt{2} - 2}\).

\(\sqrt{8 + 2 \sqrt{2}}\) and \(\sqrt{8 - 2 \sqrt{2}}\).

\(\sqrt{4 + 2\sqrt{2}}\) and \(\sqrt{4 - 2 \sqrt{2}}\).

\(\sqrt{8 + 4 \sqrt{2}}\) and \(\sqrt{8 - 4 \sqrt{2}}\).

\(\sqrt{4 \sqrt{2} + 2}\) and \(\sqrt{4 \sqrt{2} - 2}\).

(cod: P-80-8-10) Let \(a_1, a_2, \dots, a_n\) be positive numbers and consider the function \[f(t) = \left\{\begin{array}{ll} \left(\dfrac{a_1^t + \cdots + a_n^t}{n}\right)^{\frac{1}{t}}, & \text{ if } t \ne 0 \\ A, & \text{ if } t=0.\end{array} \right. \] Observe that, for each fixed \(t \ne 0\), \(f(t)\) represents a type of mean of the numbers \(a_1, \dots, a_n\). For example, \(f(1)\) is the arithmetic mean of these numbers, while \(f(-1)\) is the harmonic mean of the same numbers. Determine the value of the constant \(A\) so that the function \(f\) is continuous at \(t=0\).

\(A = \dfrac{a_1 + \cdots + a_n}{n}\)

\(A = + \infty\)

\(A = \sqrt[n]{a_1\cdots a_n}\)

\(A = \sqrt{\dfrac{a_1^2 + \cdots + a_n^2}{n}}\)

\(A = 1\)

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Date:	Wednesday, 24 June 2026, 3:53 AM

Derivatives

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