Section 4.2 Derivative Rules for Combinations of Functions
In the last section we learned rules to symbolically differentiate some elementary functions. To summarize, we have established 5 rules.
However, we do not yet have a rule for taking the derivative of a function as simple as Rather than producing rules for each kind of function, we wish to discover how to differentiate functions obtained by arithmetic on functions we already know how to differentiate. This would let us differentiate functions like or or which are built up from our elementary functions. We want rules for multiplying a known function by a constant, for adding or subtracting two known functions, and for multiplying or dividing two known functions.
Subsection 4.2.1 Derivatives of scalar products
We start by differentiating a constant times a function.
Claim 4.2.1. Scalar multiple rule.
Example 4.2.2. Derivatives of constants times standard functions.
Solution.
Using our rule:
Subsection 4.2.2 Derivatives of sums and differences
Next we want to look at the sum or difference of two functions.
Claim 4.2.3. Sum and difference rule.
Example 4.2.4. Derivatives of sums and differences of standard functions.
Solution.
Using our rule:
Theory and justification.
The basic argument for all of our rules starts with local linearity. Recall that if is differentiable at then in a region around we can approximate by a linear function, To find the derivative of a scalar product, sum, difference, product, or quotient of known functions, we perform the appropriate actions on the linear approximations of those functions. We then take the coefficient of the linear term of the result.
For our first rule we are differentiating a constant times a function. Following the general method we look at how we multiply a constant times the linear approximation.
Taking the coefficient of the linear term gives the scalar multiple rule, the derivative of a constant times a functions is the constant times the derivative of the function.
Next, we want to look at the sum or difference of two functions. Following the general method we look at the sum or difference of the linear approximations.
Taking the coefficient of the linear term gives the sum or difference rule, the derivative of a sum or difference of two functions is the sum or difference of the derivatives of the functions.
Subsection 4.2.3 Derivatives of products
We now turn our attention to the product of two functions.
Claim 4.2.5. Product rule.
Warning: Note that the derivative of a product is not the product of the derivatives!
We start with an example that we can do by multiplying before taking the derivative. This gives us a way to check that we have the rule correct.
Example 4.2.6. Simple derivative of a product.
Solution.
Note that Using our rule for monomials Using the same rule we see and We can now evaluate using the product rule:
Both methods give the same answer. Note that the product of the derivatives is which is NOT the derivative of the product.
Example 4.2.7. General derivatives of products.
Solution.
Theory and justification.
Following the general rule we look at the linear term of the product of the linear approximations. Consider the product of two linear expressions.
The coefficient of the linear term is Thus, when we take the product
the coefficient of the linear term is
Subsection 4.2.4 Derivatives of quotients
Finally, we turn our attention to the quotient of two functions.
Claim 4.2.8. Quotient rule.
Warning: Once again, note that the derivative of a quotient is NOT the quotient of the derivatives!
Example 4.2.9. Simple derivative of a quotient.
For our first example we look at a case that we dane do without the quotient rule by simplifying first. This lets us check our answer. Let and Find the derivative of
Solution.
We start by simplifying. Note that Using our rule for monomials,
Now we use the quotient rule directly. Using the same rule we see and Using the quotient rule:
Both methods give the same answer.
Note that the quotient of the derivatives is which is not the same as the derivative of the quotient.
Example 4.2.10. General derivatives of quotients.
Solution.
Theory and justification.
Following the general method, we look at the linear term of the quotient of the linear approximations. However, we need to do an algebraic trick before we can find the linear term. Consider the quotient of two linear expressions:
When is small enough, we get a good approximation by ignoring the term. In that approximation, the coefficient of the linear term is Thus, when we take the quotient,
the coefficient of the linear term is
Reading Questions 4.2.5 Reading Check
1. Reading check, Derivative Rules for Combinations of Functions.
This question checks your reading comprehension of the material is section 4.2, Derivative Rules for Combinations of Functions, of Business Calculus with Excel. Based on your reading, select all statements that are correct. There may be more than one correct answer. The statements may appear in what seems to be a random order.
- The derivative of
is - The derivative of
is - The derivative of
is - The derivative of
is - The derivative of
is - The derivative of
is - The derivative of
is - The derivative of
is - None of the above
Exercises 4.2.6 Exercises: Derivative Rules for Combinations of Functions Problems
Exercise Group.
Use the rules from the last two sections to find the derivatives of the following functions.
1.
Answer.
2.
3.
Solution.
Rewrite using exponential notation: Then take the derivative:
4.
5.
Solution.
Product Rule:
6.
7.
Solution.
Product Rule:
Note: the equation is also acceptable as a solution.
Factoring is sometimes done, but depends on the problem we are trying to solve and what we are trying to do with the derivative.
8.
9.
Solution.
If we have a product of 3 functions we need to first consider 2 of the functions as a sigle function that includes a product.
10.
11.
Solution.
Quotient Rule:
We can simplify this :
12.
13.
Solution.
Quotient Rule:
14.
15.
Solution.
Quotient Rule:
16.
17.
Solution.
Quotient Rule with embedded Product Rule:
18.
Exercise Group.
For the following problems, use the following data to find the indicated derivative.
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
f(x) | 3 | 5 | 7 | 1 | 9 | 8 | 4 | 2 | 0 | 6 |
f’(x) | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 | 9 | 8 |
g(x) | 8 | 4 | 0 | 6 | 2 | 9 | 5 | 1 | 7 | 3 |
g’(x) | 6 | 8 | 4 | 2 | 0 | 7 | 9 | 3 | 5 | 1 |
23.
The profit function at the widget factory is Find the maximum profit.
Solution.
Find the critical point: So
The maximum profit is
24.
25.
The cost function for gizmo production is for Find the equation of the line tangent to the cost function at
Solution.
For the tangent line we need a point and a slope, and once we have those we find the equation of the line.
Point: When
Slope: so at we have
The line:
So we have or in slope intercept form:
26.
The formula for the current value of a particular revenue stream is Find the equation of the line tangent to the cost function at
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