I did warn you that your instructor might ask you to derive the quotient rule the hard way. And here it is, as pointless as I believe this exercise to be. So don't bother to learn this unless your instructor is likely to make you do it on an exam (a good clue to that is if he or she does this on the board).
The limit formula for a derivative is (once again):
f(x + h) - f(x)
f'(x) = lim 4.3a-1
h > 0 h
Since we are already using the symbol, h, in 4.3-1, I will
not use functions named g(x) or h(x) in this derivation.
Instead, I will say that:
u(x)
f(x) = 4.3a-2
v(x)
where u(x) and v(x) are arbitrary functions whose
derivatives exist.
So we substitute 4.3a-2 into 4.3a-1:
u(x + h) u(x)
-
v(x + h) v(x)
f'(x) = lim 4.3a-4
h > 0 h
This is already starting to look nasty, and it's gonna get worse. Now
we make the substitutions of
u(x) + h u'(x) u(x)
-
v(x) + h v'(x) v(x)
f'(x) = lim 4.3a-5
h > 0 h
We now have the difference of fractions in the numerator, so we have
to put them over a common denominator:
(u(x)v(x) + h u'(x)v(x) ) - (u(x)v(x) + h v'(x)u(x) )
v2(x) + h v'(x)v(x)
f'(x) = lim
h > 0 h
4.3a-6
And just as we reach the limit of our tolerance for nastiness, we get a
cancellation of the u(x)v(x) terms:
h u'(x)v(x) - h v'(x)u(x)
v2(x) + h v'(x)v(x)
f'(x) = lim 4.3a-7
h > 0 h
It's still kind of nasty because we have a fraction forming the numerator
of another fraction. We can fix that by dividing h out of the
bottom denominator and out of the fraction in the numerator:
h u'(x)v(x) - h v'(x)u(x)
f'(x) = lim 4.3a-8
h > 0 h v2(x) + h2v'(x)v(x)
And if you factor out the h in the numerator, you get:
h (u'(x)v(x) - v'(x)u(x) )
f'(x) = lim 4.3a-9
h > 0 h v2(x) + h2v'(x)v(x)
At last, we have our old friend, a polynomial in h in the numerator
and a polynomial in h in the denominator. Again we apply the
rule for that from section 2.5. The lowest
power of h that both numerator and denominator share is
h1. So the h2 in the denominator
goes away in the limit, and, using the rule to take the limit, we are left with:
u'(x)v(x) - v'(x)u(x)
f'(x) =
|
With just a little application of the commutative law of multiplication and some renaming of functions, you can make 4.3a-10 look exactly like equation 4.3-7. And that concludes the derivation.
email me at hahn@netsrq.com
Exercise 1: In each of these examples you must apply the rules we have discussed so far -- mostly the quotient rule. As in previous problems, the strategy is to break the function, h(x) into its components, find the derivatives of those components, then apply the appropriate rule or rules to recombine them into h'(x).
1a) This problem is clearly a quotient of two quadratics, and we
know how to take the derivative of quadratics. Let
(x2 - 3x + 2)(2x + 2) - (x2 + 2x + 1)(2x - 3)
h'(x) =
(x2 - 3x + 2)2
You can multiply this thing out if you like (you will get some cancellation
in the numerator), but what's shown above is an answer I'd accept.
1b) Here again we have a quotient. Let
(x2 + 1)(2x - 3) - ( x(x - 3) )2x
h'(x) =
(x2 + 1)2
Again, you can multply this out if you feel you need the algebra practice,
but you shouldn't have to.
1c) This one should be easy after you learned that the rule for
taking the derivative of
h'(x) = -9x-4
1d) Again we have a quotient. We don't have an expression for
f(x), but we can set
(x3 - 4x2 + 3x - 1)f'(x) - f(x)(3x2 - 8x + 3)
h'(x) =
(x3 - 4x2 + 3x - 1)2
1e) Yet another quotient. This time we don't have an expression
for g(x), but we can set
g(x)(2x) - x2g'(x)
h'(x) =
g2(x)
1f) This one gave you two choices. You could use the rule about
negative exponents to find the derivative of the denominator, then apply
the quotient rule, or you could multipy numerator and denominator by
x2, and then apply the quotient rule to a more
traditional quotient. Here is the first way. Set
(x-2 - x-1)(0) - (1)(-2x-3 + x-2)
h'(x) =
The left half of the numerator drops out because it is multiplied by zero.
This leaves you with
(x-2 - x-1)2
2x-3 - x-2
h'(x) =
And that is a valid answer. Here is the other way. Multiplying numerator
and denominator of the original expression by x2 gives
(x-2 - x-1)2
x2
h(x) =
Now let
1 - x
(1 - x)(2x) - x2(-1)
h'(x) =
which is also a valid answer. Do you have enough algebraic fortitude to
demonstrate that the two answers I've given to this problem are equivalent?
(1 - x)2
1g) This one is easy. I threw you a curve by putting f(x) in the denominator even though the names we used in our formulation of the quotient rule had g(x) in the denominator. But what we call a function shouldn't make any difference. A denominator by any other name is still a denominator. If we use the formulation I gave in words for the quotient rule, you will be able to say that the numerator is 1 (a constant), the derivative of the numerator is therefore zero, the denominator is f(x), and the derivative of the denominator is f'(x). Applying the word description of the quotient rule you get:
f(x)(0) - (1)f'(x)
h'(x) =
f2(x)
Again, the left half of the numerator drops out because it is multiplied by
zero, leaving:
-f'(x)
h'(x) =
f2(x)
Exercise 2: The formula given by the answer to 1g is a general rule for finding the derivative of the reciprocal (which means 1 over something) of any function whose derivative exists.
Exercise 3: In this problem, both the numerator and the denominator are the same. If we apply the quotient rule to that, we get:
f(x)f'(x) - f(x)f'(x)
h'(x) =
f2(x)
Observe that the numerator cancels, giving Notice that the algebraic method of solving this gave us a value of zero for h'(x) for every x where f(x) is not zero. This method gives us a value only where f(x) is not zero and f'(x) exists. Where f'(x) does not exists (and remember that not all functions have derivatives defined everywhere, and some don't have them defined at all), the above expression for h'(x) is undefined.
email me at hahn@netsrq.com