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Calculation Thought Experiment
The calculation thought experiment is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, or difference. Given such an expression, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient, and so on. Using the Calculation Thought Experiment (CTE) to Differentiate a Function
Examples
1. (3x2-4)(2x+1) can be computed by first calculating the expressions in parentheses and then multiplying. Since the last step is multiplication, we can treat the expression as a product. 2. (2x-1)/x can be computed by first calculating the numerator and denominator, and then dividing one by the other. Since the last step is division, we can treat the expression as a quotient. 3. x2 + (4x-1)(x+2) can be computed by first calculating x2, then calculating the product (4x-1)(x+2), and finally adding the two answers. Thus, we can treat the expression as a sum. 4. (3x2-1)5 can be computed by first calculating the expression in parentheses, and then raising the answer to the fifth power. Thus, we can treat the expression as a power. |
Using the Calculation Thought Experiment (CTE)
Let us use the CTE to find the derivative of
| (3x+1) | ![]() x2+x | . |
To use this method, pretend you were calculating, one step at a time, the value of for, say, x = 5. (You don't need to actually do the calculation.) One way of doing the calculation would be to use the following procedure:
Since the last operation is multiplication, the CTE tells us that the given expression is a product and so we should use the product rule.
| f(x) = (3x+1) | ![]() x2+x |
. |
Thus,
| f'(x) | = | ![]() dx |
(3x+1) | ![]() x2+x |
+ | ![]() dx |
![]() x2+x |
. . . . (I) | |||||||
| (second left alone) + (first left alone) | (deriv. of second) | ||||||||||||||
Remember that the expressions "d/dx" are short-hand for "the derivative of ..." In other words, we haven't done the work yet; the line above is just telling us what we need to do. (If we wanted, we could take a coffee break and come back to it later to do the work.)
To finish the calculation, we must compute the magenta- and blue-colored derivatives one-at-a-time and plug them in to the expression above:
The first (magenta) derivative is easy:
![]() dx |
(3x+1) = 3 |
![]() dx |
![]() x2+x |
= | ![]() (x2+x)2 | = | (x2+x)2 |
Now substitute these derivatives into formula (I) to obtain the answer:
| f'(x) | = | 3 | ![]() x2+x |
+ | (x2+x)2 |
