- The product rule
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- As part (b) of Preview Activity shows,
- it is not true in general that the derivative of a product of two functions is the product of the derivatives of those functions.
- To see why this is the case,
- we consider an example involving meaningful functions.
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- Say that an investor is regularly purchasing stock in a particular company.
- Let N(t) represent the number of shares owned on day t,
- where t = 0 represents the first day on which shares were purchased.
- Let S(t) give the value of one share of the stock on day t;
- note that the units on S(t) are dollars per share.
- To compute the total value of the stock on day t,
- we take the product
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- V(t) = N(t) \, \text{shares} \cdot S(t) \, \text{dollars per share}
- .
- Observe that over time,
- both the number of shares and the value of a given share will vary.
- The derivative N'(t) measures the rate at which the number of shares is changing,
- while S'(t) measures the rate at which the value per share is changing.
- How do these respective rates of change affect the rate of change of the total value function?
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- To help us understand the relationship among changes in N,
- S, and V, let's consider some specific data.
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- Suppose that on day 100,
- the investor owns 520 shares of stock and the stock's current value is $27.50 per share.
- This tells us that N(100) = 520 and S(100) = 27.50.
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- On day 100,
- the investor purchases an additional 12 shares
- (so the number of shares held is rising at a rate of 12 shares per day).
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- On that same day the price of the stock is rising at a rate of 0.75 dollars per share per day.
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- In calculus notation, the latter two facts tell us
- that N'(100) = 12
- (shares per day)
- and S'(100) = 0.75
- (dollars per share per day).
- At what rate is the value of the investor's total holdings changing on day 100?
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+ Foundational Exercises
- Observe that the increase in total value comes from two sources:
- the growing number of shares,
- and the rising value of each share.
- If only the number of shares is increasing
- (and the value of each share is constant),
- the rate at which total value would rise is the product of the current value of the shares and the rate at which the number of shares is changing.
- That is, the rate at which total value would change is given by
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- S(100) \cdot N'(100) = 27.50 \, \frac{\text{dollars} }{\text{share} } \cdot 12 \, \frac{\text{shares} }{\text{day} } = 330 \, \frac{\text{dollars} }{\text{day} }
- .
+ Coming soon.
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- Note particularly how the units make sense and show the rate at which the total value V is changing,
- measured in dollars per day.
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- If instead the number of shares is constant,
- but the value of each share is rising,
- the rate at which the total value would rise is the product of the number of shares and the rate of change of share value.
- The total value is rising at a rate of
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- N(100) \cdot S'(100) = 520 \, \text{shares} \cdot 0.75 \, \frac{\text{dollars per share} }{\text{day} } = 390 \, \frac{\text{dollars} }{\text{day} }
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- Of course, when both the number of shares and the value of each share are changing,
- we have to include both of these sources.
- In that case the rate at which the total value is rising is
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- V'(100) = S(100) \cdot N'(100) + N(100) \cdot S'(100) = 330 + 390 = 720 \, \frac{\text{dollars} }{\text{day} }
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- We expect the total value of the investor's holdings to rise by about $720 on the 100th day.
- While this example highlights why the product rule is true,
- there are some subtle issues to recognize.
- For one, if the stock's value really does rise exactly $0.75 on day 100,
- and the number of shares really rises by 12 on day 100,
- then we'd expect that V(101) = N(101) \cdot S(101) = 532 \cdot 28.25 = 15029.
- If, as noted above,
- we expect the total value to rise by $720,
- then with V(100) = N(100) \cdot S(100) = 520 \cdot 27.50 = 14300,
- then it seems we should find that V(101) = V(100) + 720 = 15020.
- Why do the two results differ by 9?
- One way to understand why this difference occurs is to recognize that N'(100) = 12 represents an
- instantaneous rate of change,
- while our (informal) discussion has also thought of this number as the total change in the number of shares over the course of a single day.
- The formal proof of the product rule reconciles this issue by taking the limit as the change in the input tends to zero.
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- Next, we expand our perspective from the specific example above to the more general and abstract setting of a product P of two differentiable functions,
- f and g.
- If P(x) = f(x) \cdot g(x),
- our work above suggests that P'(x) = f(x) g'(x) + g(x) f'(x).
- Indeed, a formal proof using the limit definition of the derivative can be given to show that the following rule,
- called the product rule,
- product rule
- holds in general.
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- The Product Rule
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- product rule
- If f and g are differentiable functions,
- then their product P(x) = f(x) \cdot g(x) is also a differentiable function, and
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- P'(x) = f(x) g'(x) + g(x) f'(x)
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- In light of the earlier example involving shares of stock,
- the product rule also makes sense intuitively:
- the rate of change of P should take into account both how fast f and g are changing,
- as well as how large f and g are at the point of interest.
- In words the product rule says:
- if P is the product of two functions f
- (the first function)
- and g
- (the second),
- then the derivative of P is the first times the derivative of the second,
- plus the second times the derivative of the first.
- It is often a helpful mental exercise to say this phrasing aloud when executing the product rule.
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- Use the product rule to differentiate P(z) = z^3 \cdot \cos(z).
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- In P(z) = z^3 \cdot \cos(z),
- the first function is z^3 and the second function is \cos(z).
- By the product rule, P' will be given by the first,
- z^3,
- times the derivative of the second,
- -\sin(z), plus the second,
- \cos(z), times the derivative of the first, 3z^2.
- That is,
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- P'(z) = z^3(-\sin(z)) + \cos(z) 3z^2 = -z^3 \sin(z) + 3z^2 \cos(z)
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- The quotient rule
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- Because quotients and products are closely linked,
- we can use the product rule to understand how to take the derivative of a quotient.
- Let Q(x) be defined by Q(x) = f(x)/g(x),
- where f and g are both differentiable functions.
- It turns out that Q is differentiable everywhere that g(x) \ne 0.
- We would like a formula for Q' in terms of f,
- g, f', and g'.
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- Multiplying both sides of the formula Q = f/g by g,
- we observe that
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- f(x) = Q(x) \cdot g(x)
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- Applying the product rule to differentiate f, it follows
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- f'(x) = Q(x) g'(x) + g(x) Q'(x)
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+ Additional Practice Exercises
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- Since we want to know a formula for Q',
- we solve this most recent equation for Q'(x) and find
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- Q'(x) g(x) = f'(x) - Q(x) g'(x)
- .
- Dividing both sides by g(x), we have
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- Q'(x) = \frac{f'(x) - Q(x) g'(x)}{g(x)}
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- Finally, we recall that Q(x) = \frac{f(x)}{g(x)}.
- Substituting this expression in the preceding equation and simplifying, it follows that
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- Q'(x) \amp= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)}
- \amp= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} \cdot \frac{g(x)}{g(x)}
- \amp= \frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2}
- .
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- The preceding argument results in the
- quotient rule.
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- The Quotient Rule
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- quotient rule
- If f and g are differentiable functions,
- then their quotient Q(x) = \frac{f(x)}{g(x)} is also a differentiable function for all x where g(x) \ne 0 and
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- Q'(x) = \frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2}
- .
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- As with the product rule,
- it can be helpful to think of the quotient rule verbally.
- If a function Q is the quotient of a top function f and a bottom function g,
- then Q' is given by the bottom times the derivative of the top,
- minus the top times the derivative of the bottom,
- all over the bottom squared.
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- Use the quotient rule to differentiate Q(t) = \dfrac{\sin(t)}{2^t}.
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- Since Q(t) = \frac{\sin(t)}{2^t},
- we call \sin(t) the top function and 2^t the bottom function.
- By the quotient rule,
- Q' is given by the bottom,
- 2^t,
- times the derivative of the top, \cos(t),
- minus the top, \sin(t),
- times the derivative of the bottom,
- 2^t \ln(2), all over the bottom squared, (2^t)^2.
- That is,
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- Q'(t) = \frac{2^t \cos(t) - \sin(t) 2^t \ln(2)}{(2^t)^2}
- .
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- In this particular example,
- it is possible to simplify Q'(t) by removing a factor of 2^t from both the numerator and denominator,
- so that
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- Q'(t) = \frac{\cos(t) - \sin(t) \ln(2)}{2^t}
- .
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- In general,
- we must be careful in doing any such simplification,
- as we don't want to execute the quotient rule correctly but then make an algebra error.
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