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Calculus
Of One Real Variable – By Pheng Kim Ving |
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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
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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
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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
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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:
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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
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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
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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.
Problems & Solutions |
1. Differentiate the following functions.
Solution
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.
3. Find the tangent and normal lines to the curve y = (x + 1)/(x – 1) at x = 2.
Solution
4. Find all points on the curve y = x + (1/x) where the tangent line is horizontal.
Solution
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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