Review of Derivatives and Some Applications

Published Mar 19, 2024

Review of Limits and Derivatives

The derivative is defined as the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. The slope of a line that lies tangent to the curve at a specific point. An alternate derivative definition is the limit of the instantaneous rate of change of the function as the time between measurements decreases to zero. These are different ways to look at basically the same thing, and if these technical definitions don’t make immediate sense, don’t worry. They will be explained in more detail, with examples, throughout this tutorial.

The method of obtaining the derivative of a function is called differentiation, a process where we find the instantaneous rate of change in a function based on one of its variables. One of the most common examples is the rate change of displacement with respect to time, called velocity.

$v = dx/dt$

To conceptualize the mathematical definition of derivatives, we need a brief review of limits. In mathematics, a limit is the value that a function approaches as the input goes toward another value. Limits are fundamental to the concept of calculus and are used to define integrals, derivatives, and continuity.

Definition: The derivative of the function $f$ at $x = c$ is the limit of the slope of the secant line from $x = c$ to $x = c + h$ as $h$ approaches $0$ .

$f' \left ( x \right ) = \lim_{h \rightarrow 0} \frac{f \left ( x + h \right ) - f \left ( x \right )}{h}$

This limit is called the derivative of $f\left ( x, \right )$ at $x$ . Taking this limit is a process that creates a new function called f-prime of x (denoted by $f'\left ( x \right )$ ) from the original function $f\left ( x \right )$ .

A derivative is a rate of change, which, geometrically, is the slope of a line or the slope of the tangent line of a curve at a given point. In physics, velocity is the rate of change of position, so velocity is the derivative of position. Acceleration is the rate of change of velocity, so acceleration is the derivative of velocity. The net force is the rate of change of momentum, so the derivative of an object's momentum is the net force on the object. Suppose the electric potential is known at every point in a region of space. In that case, the electric field can be derived by the negative of the gradient (a derivative of a multi-variable function) of the electric potential. These are only a few of the applications of derivatives.

Derivative as the Slope of the Tangent

A secant is a line passing through two points of a curve (in Figure 1a, $\left ( a, f\left ( a \right ) \right )$ and $\left ( b, f\left ( b \right ) \right )$ ). As one of the points is made to approach towards the other, the secant line starts to resemble a tangent line.

Figure 1. (a) Secant line passing through points (a,f(a)) and (b,f(b))

Figure 1. (b) Visualization of how the secant line tends to look like a tangent line as P1 and P2 get closer

The slope of the tangent line at $\left ( a, f \left ( a \right ) \right )$ can be determined by the slope of the secant line that $\left ( a, f \left ( a \right ) \right )$ forms with another point on the curve $\left ( b, f\left ( b \right ) \right )$ , by the slope formula rise/run, and by using the limit relationship between the slope of a secant line and the slope of a tangent line. The difference between the y-components $f \left ( b \right ) - f \left ( a \right )$ is the ‘rise’ (or the vertical distance), and the difference between the x-components $b - a$ is the ‘run’ (or the horizontal distance). Taking the limit as $b$ approaches $a$ , we get an expression for the slope of the tangent line.

$slope \hspace{0.1cm} of \hspace{0.1cm} the \hspace{0.1cm} tangent \hspace{0.1cm} line \hspace{0.1cm} at \hspace{0.1cm} \left ( a, f\left ( a \right ) \right ) = \lim_{b \rightarrow a} \frac{f\left ( b \right ) - f\left ( a \right )}{b - a}$

To show that the slope of the tangent line is the same as the derivative of $f \left ( x \right )$ at $x$ , we define the distance between $b$ and $a$ as $h$ and write $b = a + h$ . From this equation, we can say that $b$ approaching $a$ means $h$ must approach $0$ . Substituting $b = a + h$ ,

$\lim_{b \rightarrow a} \frac{f\left ( b \right ) - f\left ( a \right )}{b - a} = \lim_{h \rightarrow 0} \frac{f\left ( a + h \right ) - f\left ( a \right )}{\left ( a + h \right ) - a} = \lim_{h \rightarrow 0} \frac{f\left ( a + h \right ) - f \left ( a \right )}{h} = f' \left ( a \right )$

We end up with the derivative of $f \left ( x \right )$ at $x = a$ or $f' \left ( a \right )$ .

Hence, the derivative of a function gives us the slope of the line tangent to the function at any point on the graph. Conversely, the slope of the tangent line at point $\left ( x, f \left ( x \right ) \right )$ is the derivative of $f \left ( x \right )$ at $x$ or $f' \left ( x \right )$ .

Derivative as the Rate of Change

Recall that the derivative is defined as the rate of change of a function with respect to a variable. An instantaneous rate of change is a limit of average rates of change as the interval approaches 0.

$\lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x} = \lim_{x_{2}\rightarrow x_{1}} \frac{y_{2}- y_{1}}{x_{2} - x_{1}} = \lim_{x_{2}\rightarrow x_{1}} \frac{f \left ( x_{2} \right ) - f \left ( x_{1} \right )}{x_{2} - x_{1}}$

We can confirm that the rate of change of f ( x ) at x is f' (x ) by considering two different values of f ( x ): f ( x₁ ) and f ( x₂ ). Again, letting x₂= x₁+ h where h is the interval between x₁ and x₂,

$\lim_{x_{2} \rightarrow x_{1}} \frac{f \left ( x_{2} \right ) - f \left ( x_{1} \right )}{x_{2} - x_{1}} = \lim_{h \rightarrow 0} \frac{f \left ( x_{1} + h \right ) - f \left ( x_{1} \right )}{\left ( x_{1} + h \right ) - x_{1}} = \lim_{h \rightarrow 0} \frac{f\left ( x_{1} + h \right ) - f\left ( x_{1} \right )}{h} = f' \left ( x_{1} \right )$

General Differentiation Formulas

Calculating derivatives using the limit definition can be time-consuming and difficult. Therefore, differentiation formulas are derived from the definition to make the process more convenient. Some important rules are the power rule, sum and difference rule, product rule, quotient rule, and chain rule.

In these equations, note that $f' \left ( x \right )$ is also denoted as $d \left ( f \left ( x \right ) \right ) / dx.$

Derivative of a constant

$\frac{d}{dx} \left ( c \right ) = 0$

Derivative of a constant multiplied with function $f$

$\frac{d}{dx} \left ( c \cdot f\left ( x \right ) \right ) = cf' \left ( x \right )$

Power Rule

$\frac{d}{dx}\left ( x^{n} \right ) = nx^{n-1}$

Sum and difference Rule

$\frac{d}{dx} \left [ f\left ( x \right ) \pm g\left ( x \right ) \right ] = f' \left ( x \right ) \pm g' \left ( x \right )$

Product Rule

$\frac{d}{dx} \left [ f \left ( x \right ) g \left ( x \right ) \right ] = f' \left ( x \right ) g \left ( x \right ) + f \left ( x \right ) g' \left ( x \right )$

Quotient Rule

$\frac{d}{dx} \left [ \frac{f\left ( x \right )}{g\left ( x \right )} \right ] = \frac{f' \left ( x \right ) g \left ( x \right ) - f \left ( x \right ) g' \left ( x \right )}{\left [g\left ( x \right ) \right ]^{2}}$

Chain Rule

If a function $y = f \left ( x \right ) = g \left ( u \right )$ and if $u = h\left ( x \right )$ , then the chain rule for differentiating the composite function $y$ is defined as

$\frac{dy}{dx} = \left ( \frac{dy}{du} \right ) \left ( \frac{du}{dx} \right )$

Differentiation formulas for trigonometric and logarithmic functions are also derived from the definition. Some of them are:

$\frac{d}{dx} \left ( sin \hspace{0.1cm} x \right ) = cos \hspace{0.1cm} x$

$\frac{d}{dx} \left ( cos \hspace{0.1cm} x \right ) = - sin \hspace{0.1cm} x$

$\frac{d}{dx} \left ( sinh \hspace{0.1cm} x\right ) = cosh \hspace{0.1cm} x$

$\frac{d}{dx} \left ( cosh \hspace{0.1cm} x \right ) = sinh \hspace{0.1cm} x$

$\frac{d}{dx}\left ( c^{f\left ( x \right )} \right ) = f' \left ( x \right ) \cdot c^{f\left ( x \right )} ln \hspace{0.1cm} c$

$\frac{d}{dx}\left ( ln \hspace{0.1cm} x\right ) = \frac{1}{x}$

$\frac{d}{dx}\left [ log_{a} \hspace{0.1cm} f\left ( x \right )\right ] = \frac{1}{ln \hspace{0.1cm} a} \cdot \frac{1}{f\left ( x \right )} \cdot f'\left ( x \right )$

Example: Given the trigonometric identity $tan \hspace{0.1cm} x = sin \hspace{0.1cm} x / cos \hspace{0.1cm} x$ determine the derivative of $tan \hspace{0.1cm} x$ .

$\begin{aligned} \frac{d}{dx}\left ( tan \hspace{0.1cm} x \right ) & = \frac{d}{dx} \left ( \frac{sin \hspace{0.1cm} x }{cos \hspace{0.1cm} x} \right ) \\ & = \frac{\frac{d}{dx} \left ( sin \hspace{0.1cm} x \right ) \cdot cos \hspace{0.1cm} x - sin \hspace{0.1cm} x \cdot \frac{d}{dx} \left ( cos \hspace{0.1cm} x \right )}{\left ( cos \hspace{0.1cm} x \right )^{2}} \hspace{1.2cm} \left ( using \hspace{0.1cm} quotient \hspace{0.1cm} rule \right )\\ & = \frac{\left ( cos \hspace{0.1cm} x \right ) \left ( cos \hspace{0.1cm} x \right ) - \left ( sin \hspace{0.1cm} x \right ) \left ( - sin \hspace{0.1cm} x \right )}{cos^{2} \hspace{0.1cm} x} \\ & = \frac{cos^{2} \hspace{0.1cm} x + sin^{2} \hspace{0.1cm} x}{cos^{2} \hspace{0.1cm} x} = \frac{1}{cos^{2} \hspace{0.1cm} x} \hspace{2.9cm} \left ( Phytagorean \hspace{0.1cm} identity \right ) \end{aligned}$

$\boxed{\frac{d}{dx} \left ( tan \hspace{0.1cm} x \right ) = sec^{2} \hspace{0.1cm} x}$

Example: Given $f \left ( x \right ) = x^{3} + 3x^{2} - 7x + 4$ , find $f' \left ( x \right )$ .

For polynomials, we take the derivative of each term using the power rule and add/subtract them accordingly using the sum and difference rule.

$f'\left ( x \right ) = \left ( 3 \cdot x^{3-1} \right ) + \left ( 2 \cdot 3 x^{2-1} \right ) - \left ( 1 \cdot 7x^{1-1} \right ) + 0$

$f'\left ( x \right ) = 3x^{2} + 6x - 7$

Example: Given $f\left ( x \right ) = \left ( 2x + 5 \right )^{4}$ , find $f'\left ( x \right )$ .

When at least two terms are in quantity raised to an exponent, the chain rule simplifies the differentiation process. Here, we see that $y = f\left ( x \right )$ can be expressed as $f\left ( x \right ) = g\left ( u \right ) = u^{4}$ where $u = 2x + 5$ .

Using the chain rule

$\frac{dy}{dx} = \left ( \frac{dy}{du} \right )\left ( \frac{du}{dx} \right )$

$dy / du = \frac{d}{du}\left ( u^{4} \right ) = 4 \cdot u^{4-1} = 4u^{3}$ and

$du / dx = \frac{d}{dx} \left ( 2x + 5 \right ) = 2 \cdot x^{1-1} + 0 = 2$

$f'\left ( x \right ) = dy / dx = \left ( dy / du \right )\left ( du / dx \right ) = \left ( 4u^{3} \right )\left ( 2 \right ) = 8u^{3}$

$f'\left ( x \right ) = 8\left ( 2x + 5 \right )^{3}$

Example: Given $f\left ( x \right ) = \left ( x^{2} - 3 \right )\left ( x^{3} + 5x - 1 \right )$ , find $f' \left ( x \right )$ .

Since the function is expressed as a product, we can use the product rule:

$f\left ( x \right ) = u\left ( x \right )v\left ( x \right ) \hspace{0.5cm} \Rightarrow \hspace{0.5cm} f'\left ( x \right ) = u'\left ( x \right )v\left ( x \right ) + u\left ( x \right )v'\left ( x \right )$

$\begin{aligned} f'\left ( x \right ) & = \frac{d}{dx} \left ( x^{2} - 3 \right ) \cdot \left ( x^{3} + 5x - 1 \right ) + \left ( x^{2} - 3 \right ) \cdot \frac{d}{dx} \left ( x^{3} + 5x - 1 \right ) \hspace{0.8cm} \left ( product \hspace{0.1cm} rule \right )\\ & = \left ( 2x \right )\left ( x^{3} + 5x - 1 \right ) + \left ( x^{2} - 3 \right )\left ( 3x^{2} + 5 \right ) \hspace{3.5cm} \left ( power \hspace{0.1cm} rule, \hspace{0.1cm} sum/difference \hspace{0.1cm} rule \right )\\ & = 2x^{4} + 10x^{2} - 2x + 3x^{4} - 9x^{2} + 5x^{2} - 15 \\ & = \[ \boxed{5x^{4} + 6x^{2} - 2x - 15}\] \end{aligned}$

Note that we will get the same result if we multiply $u\left ( x \right )v\left ( x \right )$ first and then take the derivative of a polynomial.

$\begin{aligned} f\left ( x \right ) & = \left ( x^{2} - 3 \right )\left ( x^{3} + 5x - 1 \right ) \\ & = x^{5} + 5x^{3} - x^{2} - 3x^{3} - 15x + 3 \\ & = x^{5} 2x^{3} - x^{2} - 15x + 3 \\ f' \left ( x \right ) & = \[ \boxed{5x^{4} + 6x^{2} - 2x - 15} \hspace{1.5cm} \left ( power \hspace{0.1cm} rule, \hspace{0.1cm} sum/difference \hspace{0.1cm} rule \right )\] \end{aligned}$

Rate of Change Examples

Velocity: The position of a moving object is given by $r\left ( t \right ) = 4t^{2} - 5t + 1$ in centimeters. What is the instantaneous velocity at $t = 10 \hspace{0.1cm} s$ ?

Solution: The instantaneous velocity is the first derivative of the position at a given point. For this problem,

$\begin{aligned} r\left ( t \right ) & = 4t^{2} - 5t + 1 \\ r' \left ( t \right )& = v_{ins} = 8t - 5 \\ v_{ins} & = 8\left ( 10 \right ) - 5 \\ v_{ins} & = \[ \boxed{75 \hspace{0.1cm} cm/s}\] \end{aligned}$

Function: Determine the point on the function that is not changing: $y = 2x^{2} - 12x + 9$

Solution: In order to determine where the function is not changing, it is necessary to take the derivative and set the result equal to zero. This will provide information on where the slope of the tangent line is zero, hence where the curve is not changing.

Figure 2: Examples of when the slope of the tangent line is zero. As shown, tangent lines with a slope of zero are observed at the crests (relative maxima) and troughs (relative minima) of curves. At these points, the derivative or the instantaneous rate of change of f(x) is zero.

$dr / dt = 0.5$

Once we find the x-value that gives the derivative a value of zero, we then substitute this x-value back into the original function to obtain the point.

$\begin{aligned} y & = 2x^{2} - 12x + 9 \\ y' & = 4x - 12 = 0 \\ x & = 12/4 = 3 \end{aligned}$

Substituting $x = 3$ to the original function to solve for $y$ ,

$\begin{aligned} y & = 2\left ( 3 \right )^{2} - 12\left ( 3 \right ) + 9 = 18 - 36 + 9 \\ y & = - 9 \end{aligned}$

The point where the function is not changing is (3,-9).

Solution: The volume of a sphere is given by $V = \left ( 4/3 \right )\pi r^{3}$ . Substituting the radius $r\left ( t \right ) = 5 - 6e^{-\left ( 1/3 \right )t}$ , the volume can be expressed as $V = \frac{4}{3}\pi \left ( 5 - 6e^{-\frac{1}{3}t} \right )^{3}$

We now have an expression for the volume that is dependent on $t \Rightarrow V\left ( t \right )$ . Taking the derivative with respect to $t$ , $\begin{aligned} \frac{d}{dt} \left ( V\left ( t \right ) \right )& = \frac{dV}{dt}\\ & = 3 \cdot \frac{4}{3}\pi \left ( 5 - 6e^{-\frac{1}{3}t}\right )^{2} \cdot \left ( 2e^{-\frac{1}{3}t} \right ) \hspace{1.5cm} \left ( chain \hspace{0.1cm} rule, \hspace{0.1cm} power \hspace{0.1cm} rule, \hspace{0.1cm} d \left ( e^{u} \right ) = e^{u}du \right)\\ & = 4\pi \left ( 5 - 6e^{-\frac{1}{3}} \right )^{2} \left ( 2e^{-\frac{1}{3}t} \right ) \\ \end{aligned}$

Substituting $t = ln \hspace{0.1cm} 8$ ,

$\begin{aligned} dV_{t=ln \hspace{0.1cm} 8} & = 4\pi \left ( 5 - 6e^{-\frac{1}{3}\left ( ln \hspace{0.1cm} 8 \right )} \right )^{2} \left ( 2e^{-\frac{1}{3}\left ( ln \hspace{0.1cm} 8 \right )} \right )\\ & = 4\pi \left ( 5 - 6e^{ln \hspace{0.1cm} \frac{1}{2}} \right )^{2} \left ( 2e^{ln \hspace{0.1cm} \frac{1}{2}} \right ) \\ & = 4\pi \left ( 5 - 6 \cdot \frac{1}{2} \right )^{2} \left ( 2 \cdot \frac{1}{2} \right )\\ & = \[ \boxed{16\pi }\] \end{aligned}$

Surface Area of a Cylinder: Two bases of a cylinder currently have the same radius of 3 units, though both bases begin to expand at a rate of 0.5 units. With both bases expanding while the height remains the same, what is the rate of growth of the cylinder’s surface area?

Solution: The surface area of a cylinder is given by $A = 2\pi rh + 2\pi r^{2}$ . From the given, we have $r = 3$ and $dr/dt = 0.5$ . To find the rate of change of the surface area, take the partial derivative of the area. A Partial Derivative is defined as a derivative in which some variables are kept constant (in this case $h$ ), and the derivative of a function with respect to another variable (in this case $r$ ) can be determined.

$\begin{aligned} A & = 2\pi rh + 2\pi r^{2} \\ \frac{dA}{dt} & = 2\pi h\frac{dr}{dt} + 4\pi r\frac{dr}{dt} = \left ( 2\pi h + 4\pi r \right )\frac{dr}{dt} \end{aligned}$

Substituting $r = 3$ and $dr / dt = 0.5$ ,

$\begin{aligned} \frac{dA}{dt} & = \left [ 2\pi h + 4\pi \left ( 3 \right ) \right ] \left ( 0.5 \right )\\ & = \[ \boxed{\left ( h + 6 \right ) \pi }\] \end{aligned}$

Note that this result is the instantaneous rate of change of the surface area of the cylinder at the current time when r=3 units. In general, the rate of change of the surface area of a cylinder when the radius varies while the height is constant (dA/dr) is given by

$\begin{aligned} \frac{dA}{dt} & = \left ( 2\pi h + 4\pi r \right ) \frac{dr}{dt} \\ dA & = \left ( 2\pi h + 4\pi r \right ) dr \\ \frac{dA}{dr} & = 2\pi \left ( h + 2r \right ) \end{aligned}$

If the height varies while the radius is constant, $dA / dh$ is given by

$\begin{aligned} \frac{dA}{dt} & = 2\pi r \frac{dh}{dt} + 0\\ \frac{dA}{dh} & = 2\pi r \end{aligned}$

This tells us that increasing the height of a cylinder with a constant radius can be thought of as stacking up rings of circumference $2\pi r$ and incremental height $dh: dA = 2\pi r \hspace{0.1cm} dh$ .

When both $h$ and $r$ vary,

$\begin{aligned} \frac{dA \left ( h, r \right )}{dt} & = 2\pi r \frac{dh}{dt} + 2\pi h \frac{dr}{dt} + 4\pi r \frac{dr}{dt} \\ & = 2\pi r \frac{dh}{dt} + 2\pi \left ( h + 2r \right ) \frac{dr}{dt} \\ & = \frac{dA}{dh} \cdot \frac{dh}{dt} + \frac{dA}{dr} \cdot \frac{dr}{dt} \end{aligned}$

In this tutorial, we defined, interpreted, and discussed the rate of change application of derivatives. The use of differentials in science and engineering is extensive, and developing a deep understanding of derivatives along with proficient differentiation skills is essential in solving practical problems in various fields of study. In the next tutorial, we will see that our knowledge of the application of derivatives, particularly as the rate of change in measuring geometric figures, is imperative in breaking down and exploring other concepts of electromagnetism.

Authored By

Susie Maestre

Susie is an Electronics Engineer and is currently studying Microelectronics. She loves fictional novels, motivational books as much as she loves electronics and electrical stuffs. Some of her fields of interests are digital designs, biomedical electronics, semiconductor physics, and photonics.

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