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Computing Derivatives

Techniques of Differentiation



In this section, we introduce the basic techniques of differentiation and apply them to functions built up from the elementary functions.

Basic Properties of Differentiation

There are two simple properties of differentiation that make the calculation of derivatives much easier. Let f (x) , g(x) be two functions, and let c be a constant. Then

  1. [cf (x)] = cf'(x)
  2. (f + g)'(x) = f'(x) + g'(x)
In words, these properties say that the derivative of a constant times a function is that constant times the derivative of the function, and the derivative of a sum of functions is the sum of the derivatives of the functions.

Product Rule

Given two functions f (x) , g(x) , and their derivatives f'(x) , g'(x) , we would like to be able to calculate the derivative of the product function f (x)g(x) . We do this by follwowing the product rule:

[f (x)g(x)] =  
  = +  
  = f (x + ε) g(x)  
  = f (x)g'(x) + g(x)f'(x)  

Quotient Rule

Now we show how to express the derivative of the quotient of two functions f (x) , g(x) in terms of their derivatives f'(x) , g'(x) . Let q(x) = f (x)/g(x) . Then f (x) = q(x)g(x) , so by the product rule, f'(x) = q(x)g'(x) + g(x)q'(x) . Solving for q'(x) , we obtain

q'(x) = = =    

This is known as the quotient rule. As an example of the use of the quotient rule, consider the rational function q(x) = x/(x + 1) . Here f (x) = x and g(x) = x + 1 , so

q'(x) = = =    

Chain Rule

Suppose a function h is a composition of two other functions, that is, h(x) = f (g(x)) . We would like to express the derivative of h in terms of the derivatives of f and g . To do so, follow the chain rule, given below:

h'(x) = f'(g(x))g'(x)    

Alternately, if we let y = g(x) , z = f (y) , then we may write the formula in the following way (using the alternate notation for derivatives):


This is easy to remember, because it looks like the dy are quantities that cancel. While convenient, one must be careful to realize that dy is just a notational device; it does not represent a number and cannot be haphazardly manipulated as such.

Implicit Differentiation

Sometimes we encounter an equation relating two variables that does not come from a function. One familiar example is the equation for a unit circle, x 2 + y 2 = 1 . While this equation is not a function in itself, its graph of its solutions is made up of the graph of two functions defined on the interval [- 1, 1] : f (x) = and g(x) = - . These functions are said to be implicit functions for the equation.

In the case of the unit circle, we were able to write down the implicit functions explicitly, but this is not always possible. As an example, consider the equation x 2 y 2 = x + y , the graph of whose solutions resembles an "infinite boomerang," displayed below.

Figure %: Plot of x 2 y 2 = x + y

It is not possible to find a simple formula for x or y , so we cannot write down the implicit functions. But we still may want to know the slope of the graph at a particular point, that is, the derivative of an implicit function at that point. Implicit differentiation allows us to do this.

The idea is to differentiate both sides of the equation with respect to x (using the chain rule where necessary). The two sides must remain equal under this differentiation. Then we solve for y'(x) in terms of x and y . The fact that we need to know both the x - and y -coordinates of a point in order to compute the derivative should come as no surprise, since two different points on the graph may very well have the same x - coordinate. The full set of solutions to an equation is not, in general, the graph of a function.

For the infinite boomerang, we obtain:

[x 2 y 2] = [x + y]  
x 2(2yy') + y 2(2x) = 1 + y'  
y'(2x 2 y - 1) = 1 - 2xy 2  
y' =  

Therefore, at the point (0, 0) , the slope of the graph is -1 . Note that we cannot just plug any point we like into this formula--the point must be a solution to the original equation in order for the answer to make sense.

Differentiation of Inverse Functions

We can put the chain rule and implicit differentiation to work to find the derivative of an inverse function when we already know the derivative of the function itself. Suppose we are given a function f (x) with derivative f'(x) and let g(x) be its inverse, so that g(f (x)) = f (g(x)) = x . Differentiating both sides of f (g(x)) = x , we obtain:

f'(g(x))g'(x) = 1  
g'(x) =  

Let us use this technique to find the derivative of the inverse sine function, f (x) = sin-1(x) , defined on the interval [- 1, 1] and taking values in [- Π/2, Π/2] . Since f'(x) = cos(x) , the formula tells us that g'(x) = 1/cos(sin-1(x)) = 1/ . The derivatives of the other inverse trigonometric functions are as follows:

cos(x) =  
tan(x) =  

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