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*}