Derivative (mathematics)

A derivative is the result of performing the process of differentiation. In general, the definition of a derivative dy/dx involves making small changes δx and δy to x and y and seeing what happens to the ratio δy/δx as δx andδy get smaller and smaller. The limiting value of δx/δy as δx and δy get smaller and smaller is the derivative of x with respect to y. It is denoted by dx/dy.

Overview

During the 1630’s and 1640’s researchers such as Gille de Roberval and Pierre de Fermat introduced the concept of differentiation. It only really caught on, however, when Isaace Newton and Gottfried Wilhelm von Leibniz rigorously defined their methods. Newton and Liebniz were the first to recognize the usefulness of differentiation as a general process. Liebniz derived the theory independently of Newton. Both approached differentiation with different notations and different methodologies. The two spent the latter part of their lives arguing over who was responsible for first discovering differentiation and integration and accusing each other of plagiarism. Leibniz’ notation most closely resembles that used in modern calculus. Though Newton independently arrived at the same conclusions, his path to discovery is slightly less accessible to the modern reader.

Derivative of a Function at a Point

The slope of the tangent line to the plot of a function at a given point is called the derivative of the function at that point. The derivative of a function f at x = a is

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provided the limit exists.

The Chain Rule

A rule for differentiating compositions of functions is called the chain rule. Let the derivative of a function h(x) be denoted by D{h(x)} or h'(x)..., which are alternative ways of stating Differentiate the function h with respect to x. The chain rule states formally D{f(g(x))} = f′(g(x)) g′(x).

More intuitively, think of the functions f and g as layers of the problem. Function f is the outer layer and function g is the inner layer. Thus, the chain rule tells us to first differentiate the outer layer, leaving the inner layer unchanged (the term f'( g(x))) , then differentiate the inner layer (the term g'(x)).

For example, take y=(3x + 1)2. The outer layer is the square and the inner layer is (3x + 1). Differentiate the square first, leaving (3x + 1) unchanged. Then differentiate (3x + 1). ) Thus,

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Derivative of Parametrically Defined Functions

With some functions it is not possible to express one variable (x) in terms of the other (y). Both x and y, however, can be expressed in terms of a parameter (t).

For example, consider the functions: x=cos2t and y=sin3t. Using radians, when t = 0, t = π..., x = 1, y = 0 while when t =π/2, x = y = −1, whereas when t = 3π/2, x = −1, y =1. Although it is not possible to express y as a function of x, it can be seen that, as x varies, y varies so y is a function of x. When x and y are given parametrically as functions of t, the first derivative dydx be found as a function of t according to dy/dx = dy/dt/dx/dt.

For example, for the functions x = cos2t, y = sin3t, find the first derivative dydx as a function of t and specifically at t = π6, π4, and π2.

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It is possible to find the second derivative of a function expressed in parametric form. The process is to find the first derivative of a function as above and then to differentiate again, that is, by treating the dy/dx as a new function and using the same process again.

Applications of Derivatives

The derivative of a function has many applications to problems in calculus. It may be used in curve sketching; solving maximum and minimum problems; solving distance; velocity and acceleration problems; solving related rate problems; and approximating function values.

The derivative of a function at a point is the slope of the tangent line at this point. The normal line is defined as the line that is perpendicular to the tangent line at the point of tangency.

For example, find the equation of the tangent line to the graph of at the point (−1,2).

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At the point (−1,2) , f′(−1) = −1/2 and the equation of the line is

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The motion of an object along a straight line can be modeled using calculus. Suppose an object is moving vertically along a line with a specified origin, and the position of the object at time t is represented by the position function s(t), which gives the distance of the object from the origin. The velocity of the object at time t is then just the rate of change of the position with respect to time, or v(t) = s′(t).

Similarly, the acceleration of the object at time t is the derivative of the velocity and is the second derivative of the position function, a(t) = v′(t) = s″(t). Imagine a ball being tossed up vertically: t in seconds and s in meters measuring the height above the ground (s = 0 is the ground). The position function in this case is given by s(t) = −4.9t2 + 20t and the velocity is the derivative of this, or v(t) =−9.8t + 20 (measured in meters/second). The acceleration is the derivative of the velocity, or a(t) = –9.8. This constant acceleration is due to the earth's gravity and is measured in meters/second².

To find when the ball is at ground level, solve the following equation: 0 = −4.9t2 + 20t for t. One solution is obviously t = 0, because the object started out at ground level. The other can be found using algebra and is t ≈ 4.082 seconds. So intuitively divide by 2 to find the time at the peak: 2.041 seconds.

Derivatives of Functions

Derivatives for trigonomic and logarithmic functions are beyond the scope of this article. However, derivatives of functions can be briefly discussed for algebraic and expenential functions.

An algebraic function is any function that can be built from the identity function y = x by forming linear combinations, products, quotients, and fractional powers. For example, all of the following are algebraic functions: f(x) = sqrt(x) = x1/2; g(x) = |x| = sqrt(x2); and h(x) = sqrt(|x|) = sqrt(sqrt(x2)).

The functions y = 2x and y = 3x are exponential functions. Because 20 and 30 are each 1, both plots go through the point (0,1). At the point (0,1) the gradient of y = 2x is approximately 0.69, and the gradient of y = 3x is approximately 1.1. These gradients represent the derivatives of each function. Somewhere between 2 and 3 is a value for a for which the gradient y = ax at the point (0,1) is equal to 1. It should be nearer to 3 because 1.1 is nearer to 1 than is 0.69. The required number has the value denoted by e of 2.7182818.... The derivative of ex is ex. This is a very important finding in the field of mathematics.

Bibliography

Fernandez, Oscar E. Everyday Calculus: Discovering the Hidden Math All Around Us. Princetown UP, 2014.

MacLane, Saunders. Mathematical Form and Function. New York: Springer, 2011.

Rassias, Themistocles M., ed. Handbook of Functional Equations: Functional Inequalities. New York: Springer, 2014.