Chapter 6: Rules For Finding Derivatives

Linearity of the Derivative

This means two things:

First:

If you know the derivative of a function , then the
derivative of a constant multiple is simply . i.e.

The derivative of a constant c times a function is equal to c times the derivative of the function :

= c f(x).

This principle is obvious if you think of the derivative as a rate of change. For example, it says that if f(t) is the distance function for a bicycle, and if the distance function for a car is always 5 times the distance for the bicycle, then the car must be going at 5 times the velocity of the bicycle:

5 f(t) = 5

Second :

If you know the derivatives and of two functions f(x) and g(x), then the derivative of the sum f(x)+g(x) is simply .

The derivative of the sum of two functions is equal to the sum of the derivatives of the functions :

( ) = f(x) + g(x).

This principle is also obvious when one thinks of rates. It says that the rate at which the sum of two distances is changing is equal to the sum of the rates at which each distance is changing.

If you have already caught on to the idea of the derivative as a limit of the difference quotient , then these are easy to understand. If y is replaced by then clearly the quotient always gets multiplied by the same constant c and so does the derivative which is its limit! Similarly, if is a sum of two functions, then their difference quotients clearly add up and give the sum formula.

+...+ .

We have:

( ) = 80 - 4.9 = = .

( ) = x^2- = = .

( ) = + ( ) = .

( ) = 1-10

= 0 .

The Product Rule

u(x+h)= (approx) ; = (approx ) .

We now look at the difference quotient for , whose limit as h approaches zero is the derivative :

= (approx)

= = .

As h approaches zero, the last term drops out, and we are left with . Thus,

if then .

This rule can be written in shorthand as follows:

Product Rule: = .

In words, the derivative of a product is equal to the first function times the derivative of the second function plus the derivative of the first function times the second function .

The above calculations certainly motivate the formula and do give an idea of the proof. They still lack many details for a complete proof. You should think about what is possibly wrong with them. The details will be discussed in class.

WARNING : Never take the derivative of a product by multiplying
the derivatives of the terms. With addition you have , but with multiplication is NOT . You should experiment with some examples and see if you can find two special functions u and v for which this is actually the case. It does work if one of u and v is the zero function or both are constant - see if you can find an example where neither is zero and at least one is not constant.

code for above diagram

Use the product rule to show how the formula can be obtained for n = 4,5,6, ... (i.e., the successive values of n after the values
n = 0,1,2,3 for which we already derived the formula earlier).


So suppose we don't yet know a formula for the derivative of , but we do know the derivative of (namely, ). Then we can write , so that, by the product rule,

= ( ) = + ( ) = .

This confirms the rule when n = 4. Now that we have established
the rule for the derivative of , we can use the same method to treat :

= ( ) = + ( ) = .

And similarly for n = 6:

= ( ) = + ( ) =

By now it should be clear that, continuing in this way, we can establish
the validity of the rule for all positive integers n. You should write out the derivation for this rule in your notes.

The Quotient Rule

Suppose that we want to find the derivative of a function that is written as the quotient of two functions whose derivatives we know: f(x) = u(x)/v(x). The rule is:

( Quotient Rule ) = ,

or, in shorthand notation,

.

This rule is a little more complicated than the product rule:

the derivative of a quotient is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all over the square of the denominator .

However it its easily understood from the product rule. Suppose we let and seek . We observe that = . Since these are equal their derivatives are equal so we have:

=

Applying the product rule to the left side we have

Since f is we then have which we can solve for to get .

As we noted above such calculations certainly motivate the formula and do give an idea of the proof. They still lack many details for a complete proof. The basic problem is that some functions don't have a derivative and in principle could be one of them. These calculations assume that actually exists.

We will derive the quotient rule again in this section, when we cover the chain rule.

=
.

The last expression can be simplified by multiplying the top and bottom by . After a little cancellation, we arrive at .

Alternative Method: The above calculations can be simplified by using proper notations and some clever simplification. First set and note that . Applying the quotient rule and linearity we get (y')' = - = .

Noticing that , we get that = .
This idea will be developed into implicit differentiation method later.

( ) =

= .

Alternately, we could use the product rule if we first write our function in the form :

( ) = = ,

which is algebraically equivalent to the answer we got using the quotient rule.

Alternative method: As above, the work can be cleaned up by using a few names. Let and . Then = . Also, . The derivative of the function becomes:

= .

Getting the final answer is now the matter of plugging in the values of y,z and simplifying.

( A bit of advice on avoiding the quotient rule .) Do not use the quotient rule when the numerator is a constant, because it is unnecessarily cumbersome in that case. For example, to find the derivative of , write this as 1 and use the power rule: the derivative is . On the other hand it is a good idea to try your quotient rule on , whose derivative is known to be - . It helps you to remember that the negative term in the numerator is .

The Second Derivative and Acceleration

As mentioned before, the second derivative is simply the derivative of the
derivative. If is our function, then there are various ways to denote the second derivative

( ):

= = = ( ).

(a) We have = ( ) = .

Then (b) is the special case , i.e., the second derivative of is .
In part (c) we have

( ) = ( ) = .

v(t) = .

The derivative of that --- the instantaneous rate at which the velocity is changing --- is what is called the acceleration often denoted a(t)

s(t)

Earlier we saw that the velocity at a certain instant t is the slope of the tangent line to the graph of the distance function at the point ( ). Similarly, the acceleration at a certain instant t is the slope of the tangent line to the graph of the velocity function at the point ( ).

First v(t) = ( ) = . Next, = = ( ) = . The distance function tells us that the bicycle is at meters from the reference point at time and is moving to the right at a certain velocity . The formula for this velocity function says that the speed of the bicycle is 5 m/sec at first (t = 0), but it increases steadily. The rate at which the velocity increases is the acceleration function , which happens to be a constant function in this example. The acceleration is measured in ``meters per second per second'' or ``meters per second squared,'' written 2 . Here are the graphs of the distance, velocity, and acceleration functions:

code for diagrams

Taking derivatives gives the formula for the velocity

and taking one more derivative gives the acceleration

The Chain Rule

and implies that =

and implies that = .

The chain rule enables us to find once we know the derivative of the outside function and the derivative of the inside function .

We shall derive the chain rule using the tangent line approximation. The t angent line approximation formula for the function can be written in the form = (approx) . Similarly, the tangent line approximation formula for the function can be written = (approx) . Putting these two facts together, we see that a small change produces a change times as much in u, which, in turn, produces a change times as much in , i.e., = (approx) = (approx) . Since we also have = (approx) , it follows that the two ``proportionality factors'' between and must be equal, i.e., . This is the chain rule. It is especially easy to remember when it is written in the notation:

( Chain Rule ) = .

It is as if the `` ,'' `` ,'' and `` '' can be treated individually as algebraic quantities, where we cancel the two `` '' on the right.

Indeed, another way of formulating the above calculations is to write

=

which is precise. Now we take the limits as goes to . We have to see why also goes to zero and then the formula will be proved. We leave the precise treatment for an appendix.

Schematically, we can regard a function of a function as a 2-step procedure, first going from x to u and then from u to y. The chain rule says that the derivative from x to u must be multiplied by the derivative from u to y to get the derivative from x to y.

Before giving some computational examples, we shall give a story problem to
show that the chain rule makes sense from a practical point of view.

Let us use to denote income and to denote the income tax. Recall that `` % tax bracket'' means that as a function of has slope over the range of values of (your income) in question. In this problem depends on , which in turn depends on the time . We could write that and together give y = . (Thus, is playing the rule of in the chain rule, and is playing the role of .) The chain rule says that

= = = .

That is, tax is increasing at the rate of $ 0.28 for every dollar increase in income, and income is increasing at the rate of $ for every year increase in time; hence, tax is increasing at the rate = dollars every year.

(a) We take = for the outside function and for the inside function. Then, since , we have

= = = = ,

as expected.

(b) Here the outside function is y = , and the inside function is . So

=

(the answer should be written in terms of x, i.e., in you should substitute , which in this case is ).

(c) Here the outside function is y = , and the inside function is . So

( )= .

When the outside function g(u) is the power function , we can write the chain rule in the form

( ) = .

For example, the derivative of the square of a function is .

(a) In the above formula for the derivative of , we take , so that . Then = .

(b) We write , and then ( ) =
= , the same answer as in part (a).

We next use the chain rule to show how we get the quotient rule. We write the quotient in the form of a product . Then we apply the chain rule to find the derivative of , where the outside function is the reciprocal function and the inside function is v(x). According to the chain rule,

=

We are now ready to apply the product rule:

= = + = = + = ,

which is the quotient rule.
Another comment to make about the chain rule is that it can be used to find the derivative of a function of a function of a function, or even a function of a function of a function of a function, and so on, where in each case we build up a function by applying a chain of simple functions.
Here is an example to illustrate this.

The outermost function is , where . In the chain rule we need , which in this case we find by using the chain rule again:

= ( ) = .

Finally,

= = = = .