How to Calculate Derivatives

Let me start with a brief historical introduction. Two men, Sir Isaac Newton and Gottfried Leibniz, independently of one another, were greatly responsible in developing the ideas surrounding the derivative.

The problem of the derivative first arouse when a mathematician, by the name of Pierre de Fermat, undertook the responsibility of determining the maxima and minima of some rather special functions. His idea was rather simple. Say you wanted to find the direction a point on the curve has (assuming that each point on the curve has a defined direction) that can be described by a tangent line. How would you go about finding the equation of the line at any specific point on the curve?

Fermat initially realized that at some points of the curve (namely the maxima and minima (note that the maxima and minima are locations on the curve that are, respectively, the highest and lowest points of the function)), the tangent lines would have a slope of zero and be directly parallel to the x-axis. He therefore had a problem. In order to find the values of the maxima and minima, he would first have to discover the equations governing the horizontal tangents (in other words, he would have to find the value of y at that point).

As Fermat reviewed his work, he realized that it raised some more general questions. For example, how would one go about determining the direction of a tangent line at any arbitrary point on the curve of the function? The attempt to find the solution to this problem is what led Pierre de Fermat to discover some of the basic principles surrounding the idea of the derivative.

Although, as stated above, the derivative was originally formulated to solve the problem regarding tangents, it was later discovered that the derivative could also shed light on finding the instantaneous rate of change at a given time for the distance function, thus calculating the instantaneous velocity at the given time. Generally speaking, the derivative could be used to calculate the rate of change of any function (well, almost any function).

Say that a bullet was fired out the barrel of an M4 Carbine rifle vertically into the air. We enact a variety of experiments and determine the muzzle velocity (initial velocity) to be 2900 fp/s (around 884 m/s). Removing air friction and other forces other than gravity from the equation, we expect the bullet to move up to a maximum point, stop momentarily, and fall back down to earth. This occurs over some time t.

Let h(t) determine the function that relates the height of the bullet in feet (also in meters for the rest of the world, there will be two separate equations) to the amount of time that has passed in seconds after firing the rifle. We know that if the force of gravity were not acting on the bullet, it would continue to travel at 2900 fp/s indefinitely.

This would mean that the bullet would travel a distance of 2900 feet for every second that passed or h(t)= 2900t. The metric system equation for this would be h(t)= 884t.

As stated above, this would not occur due to gravity. More experiments are done and the formula that includes the force of gravity is determined. It is stated as follows.

(Imperial System) h(t)= 2900t - 16t².

(Metric System) h(t) = 884t - (9.81/2)t².

The terms -16t² and - (9.81/2)t² are included in the equation due to the influence of gravity on Earth. The first term is in ft/s and the second is in m/s. Don't forget, h(t), is the value of the height in either feet or meters depending on which equation you use. Be consistent!

Note that, for the above equations, h(t) = 0 when t=0 and t=181.25s

The formula above were derived from well know physics equations. The equation used to to develop the formula is;

d(t)= v_it + (1/2)at²,

where d(t) is the distance traveled in feet (or meters) in a set time value, t, measured in seconds, v_i is the initial velocity of the fired projectile in fp/s (or m/s) and a is the acceleration in fp/s² (or m/s²). Our equations have a negative value for our "at²" expression due to the fact that gravity is labeled as a downward force and given a negative value. It is important to incorporate direction when dealing with vectors.

What the above time intervals of 0 and 181.25s state is that our bullet will be in the air for a total time of 181.25 seconds. It is important for you to understand that this formula is valid only for this time interval. This is because we are measuring the height relative to the surface of the Earth. Once the bullet hits the surface, it has returned to it's origin and will therefore have a "relative" height of 0. If it could somehow move beyond the surface and deeper into the crust, it would take on a negative value relative to the surface (because we designate up as positive and down as negative).

On to the problem. Say we wanted to find the bullets velocity at any given instant of time within our set interval limits. How would we go about doing that? Well, from past experience, we know how to calculate average velocity, so why not start there? To do so, we incorporate the idea of the average velocity during a time interval, from say, t to t+j. This is defined as the difference quotient and is expressed as follows;

This number is a number which can be calculated whenever we have both t and t+j within our set interval of [0,181.25]. The number j may be positive or negative but never zero. You can think of the value of t+j as a simple extension of t in any direction. If we were to plot it on the number line, it would look like this.

The books next to this text by Tom Apostol are highly recommended! Some of the best Calculus books I have ever used.

Keeping t static (unchanging) and taking values of j with decreasing absolute value, something interesting happens to our difference quotient. Let's take a look. Say we were to start at t=5s. If we wanted to find the average velocity of the bullet from 5s to 10s of elapsed time, we would enter these variables into our difference quotient equation and it would produce a value.

Doing the same for the time interval between 5s and 8s, it produces another value. Not that we can view both 10s and 8s as extensions of 5s. For example, 10s is simply 5s+5s and 8s is 5s+3s. Now we can see how our j is decreasing and approaching 0. Note that 10s and 8s are our t+j values. Letting j approach zero brings our interval to smaller and smaller values.

The smaller this interval becomes, the more accurate that our representation of the instantaneous velocity becomes. This is due to the fact that, at some point, the difference quotient approaches some specific value as our interval becomes smaller and smaller. This is the value of our instantaneous velocity at t=5s. See below for a visual representation of what is being said here.

As you can see, what we are doing is successively creating a smaller and smaller interval from t to t+j. When we take the slop of the secant lines between such intervals, we get average velocities. If we were to continue bringing these interval closer and closer together, as we said above, we would be reaching a limiting value where we could assume the value of the instantaneous velocity.

This could be interpreted as the limit of the secant lines as they approach the slope at point t. Of course, we can never reach the limit, but we can see the value that it approaches as j gets nearer and nearer to zero. Note that the limit here is not t, but rather, the value that the slope of the secant lines approaches as j approaches zero. We compute this through the difference quotient. The slopes of the secant lines in the above examples represent average velocities.

This can be checked by using the formula for slope. In essence, what this is trying to show is that, if we were to shorten the time interval in which we measure the average velocity to almost zero, we could get an intuitive sense as to what the instantaneous velocity would be at time t by seeing what value the limit takes on as j (the time interval) approaches zero.

This method of taking the limiting value of a difference quotient is known as calculating the derivative. It is also known as taking the limit as j approaches zero.

Below, we will introduce a step by step formula to generalize the method used above for determining the derivative of any function.

The first thing you need in order to be able to take a derivative is a function. There are some properties this function must have. This function must be defined on some open interval (a,b) on the x-axis. We then choose a fixed point x within this interval and incorporate the difference quotient.

where the number h, which may be positive or negative (but never zero), is such that x+h also lies within our interval (a,b). The numerator in the difference quotient measures the change in the function when x varies from x to x+h. The quotient itself can be referred to as the average rate of change of f in the interval from x to x+h.

Once we have found our difference quotient, we let h approach zero and see what occurs to our quotient. So long as our quotient (yes the quotient itself is a number, don't forget) approaches some limiting value L (and as long as this limiting value is the same whether h approaches zero from positive or negative values), then this limit is called the derivative of f at x and is denoted by the symbol f (prime) at x. I will introduce the symbol below. The definition of the derivative then becomes;

provided the limit exists. The number f(prime) at x is also known as the rate of change of f at x.

First, we were given a function. Taking the derivative of this function was rather simple as we had no problem with a denominator.

The second step was to take the limit of the difference quotient of the function, f,as h approaches zero. It should be obvious that the value of the function at f(x+h) is equivalent to 8(x+h)² since f(x)=8x². We simply input these values (remember that these are just numbers, nothing scary) into the difference quotient.

The next step is straightforward. We have to find a way to simplify the difference quotient so that the denominator h is not zero (we have to remove h from the denominator otherwise we cannot get a value as h approaches zero since zero cannot be in the denominator. It is undefined at zero). We do this by expanding anything we can and then simplifying.

Once we have removed h from the denominator, we are left with an expression containing h. It should be obvious that if we let h become zero in the above expression, then we are simply left with 16x. This is the value of our derivative. Plug in any x value (so long as it is within the domain of X) and the derivative will give you a corresponding value of f(prime) at x that is the rate of change of the function f(x) at x. There you have it!

Note that the process listed above of finding the derivative of a function is rather laborious and time consuming. Fortunately, there are differentiation laws to shorten this time consuming task. They can be found here. (Note that dy/dx is read "the derivative of y with respect to x and is essentially the same notation as f(prime) at x.).

The above differentiation and limit laws may use different variables than I have (for example, a instead of x) but they tell the same story. Do not be frightened by the different notation.

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