3.2 Differentiation Of Products And Quotients
| Calculus Of One Real Variable – By Pheng Kim Ving |
| 3.2 |
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| 1. Differentiation Of Products Of Functions |
Recall from algebra that the product of functions f and g is the function denoted by fg and defined by ( fg)(x) =
f(x)g(x). The value of the product fg at x is the product of the values of f and g at x. We want to find the derivative of
fg. Let’s consider an example. Let f(x) = 2x and g(x) = x3. So ( fg)(x) = (2x)(x3) = 2x4. We have f ‘(x) = 2, g‘(x) =
3x2, and ( fg)'(x) = 8x3. Now, f ‘(x)g‘(x) = (2)(3x2) = 6x2. Thus, ( fg)'(x) is not equal to f ‘(x)g‘(x).
We’ll see in the theorem below that the derivative ( fg)’ of fg is f ‘g + fg‘, not f ‘g‘! The derivative of the product is not
the product of the derivatives!
Theorem 1.1 – The Product Rule
| If f and g are differentiable, then fg is also differentiable, and: ( fg)’ = f ‘g + fg‘. |
Proof
Let x be an arbitrary point where both f and g are differentiable. Then:
EOP
Recall from Section 2.4 Theorem 1.1 that if a function is differentiable at a point, then it’s continuous there. Note also that
The General Product Rule
The product rule can be extended to more than two factors. If f, g, and h are differentiable, then fgh is also differentiable,
and:
( fgh)’ = f ‘( gh) + f( gh)’ = f ‘gh + f( g‘h + gh‘ ) = f ‘gh + fg‘h + fgh‘.
In general, if f1, f2, …, fn are differentiable, then f1 f2 …fn is also differentiable, and:
( f1 f2…fn)’ = f1‘f2…fn + f1 f2‘…fn + … + f1 f2…fn‘.
Example 1.1
Find the derivative of y = (x2 + 1)(x3 – 2).
Solution
y‘ = 2x(x3 – 2) + (x2 + 1)(3x2) = 2x4 – 4x + 3x4 + 3x2 = 5x4 + 3x2 – 4x.
EOS
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| 2. Differentiation Of The Square Root Function |
Corollary 2.1
| We have:
for all x > 0. |
Proof
Using the product rule we get:
EOP
Example 2.1
Solution
EOS
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| 3. Differentiation Of Reciprocals Of Functions |
Theorem 3.1 – The Reciprocal Rule
Proof
EOP
At any point x where f(x) = 0, f is differentiable, but 1/f isn’t, because 1/f isn’t defined there.
A Special Case
A special case is the derivative of 1/x. Since (d/dx) x = 1 we have:
Example 3.1
Find:
Solution
EOS
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| 4. The Power Rule – The Derivative Of xn |
In Section 3.1 Theorem 4.1 we see that (d/dx) xn = nxn–1 for all positive integer n. We now extend this formula to all
integers.
Corollary 4.1 – The Power Rule
| We have: for all integer n. |
Proof
EOP
We’ll see in Section 6.3 Eq. [4.1] that (d/dx) xa = axa–1 for any real number a. That’s called the general power rule.
Example 4.1
Find the derivative of f(t) = 2t3 + 4t–3.
Solution 1
f ‘(t) = 6t2 – 12t–4.
EOS
In this solution we use the power rule on both terms 2t3 and 4t–3 of f(t). The reciprocal rule can also be utilized on 4t–3,
as done in Solution 2 below.
Solution 2
EOS
We see that the power rule is much simpler than the reciprocal rule.
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| 5. Differentiation Of Quotients Of Functions |
Theorem 5.1 – The Quotient Rule
Proof
EOP
Example 5.1
Evaluate:
Solution
EOS
Remark that we substitute the value x = 2 without first simplifying. It’s simpler to do numerical calculations than to handle
algebraic symbols like the letter x.
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1. Differentiate the following functions.
Solution
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2. Evaluate:
Solution
Note
We evaluate the derivative immediately after it’s calculated, before any simplification takes place. That’s because it’s
easier to simplify an expression with numbers than with algebraic symbols.
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3. Find the tangent and normal lines to the curve y = (x + 1)/(x – 1) at x = 2.
Solution
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4. Find all points on the curve y = x + (1/x) where the tangent line is horizontal.
Solution
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5. Let b be a non-zero constant. Find the line that passes thru the point (0, b) and is tangent to the curve y = 1/x.
Solution
Suppose the line is tangent to the curve at x = a. Then the slope of the line is:
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