Box 4.2a: Answers to ExercisesKCT logo

© 1996 by Karl Hahn

1) In the first expression we have the sum of two functions, each of which can be expressed as xn, one where  n = 4  and one where  n = 3. We already know how to find the derivative of each of the summands. The sum rule says we can consider the summands separately, then add the derivatives of each to get the derivative of the sum. So the answer for the first expression is:  f'(x) = 4x3 + 3x2

In the second expression we have the sum of an xn and an  mx + b. In the first summand,  n = 2. In the second summand,  m = -7  and  b = 12. Again the sum rule says we can consider the two summands separately, then add their derivatives to get the derivative of the sum. So the answer for the second expression is:  f'(x) = 2x - 7

2) This one should have been easy. The text of the problem observed that a constant function is just a straight line function with a slope of zero (i.e.,  m = 0). We already know that the derivative of any straight line function is exactly its slope, m. When the slope is zero, so must be the derivative.

3) A very quick argument for the first part is that:

   g(x)  =  n f(x)  =  f(x) + f(x) + ... [n times] ... + f(x)
Hence, by the sum rule,
   g'(x)  =  f'(x) + f'(x) + ... [n times] ... + f'(x)  =  n f'(x)
A purist, however, would do it by induction. To get onto the first rung of the ladder, we observe that for  n = 1:
   g(x)  =  (1) f(x)  =  f(x)
   g'(x)  =  f'(x)  =  (1) f'(x)
So it works for  n = 1. Now we demonstrate that if it's true for the nth rung, then it must be true for the n + 1st rung. So if:
   gn'(x)  =  n f'(x)
   gn(x)  =  n f(x)
   gn+1(x)  =  (n + 1) f(x)  =  (n f(x) ) + f(x)
That the derivative of  n f(x)  is  n f'(x)  is given by assumption. And we know by the sum rule that:
   gn+1'(x)  =  (n f'(x) ) + f'(x)  =  (n + 1) f'(x)
And you have it proved by induction. But you only need to do it that way if you are performing for a stickler on formality.

As for the second part, you can observe that:

   n u(x)  =  f(x)
We know from the first part of the problem that:
   n u'(x)  =  f'(x)
Simply divide both sides by n, and your proof is complete.

Now, using the results from both of the parts of the problem, can you show that if:

   u(x)  =  (n/m) f(x)
and both n and m are counting numbers then:
   u'(x)  =  (n/m) f'(x)

And using an argument similar to the one we used for the second part of this problem, can you prove that the derivative of the difference is always equal to the difference of the derivatives? Hint: if  u(x) = f(x) - g(x), then  u(x) + g(x) = f(x).

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Exercise 4: In each of the functions given, you must try to break the expression up into products and sums, find the derivatives of the simpler functions, then combine them according to the sum and product rules and their derivatives.

4a) Here you are asked to find the derivative of the product of two straight line functions (both of which are in the standard  mx + b  form). So let  f(x) = m1x + b1, and let  g(x) = m2x + b2. Both are f(x) and g(x) are straight line functions. And we know that the derivative of a straight line function is always the slope of the line. Hence we know that  f'(x) = m1  and  g'(x) = m2. Now simply apply the product rule (equation 4.2-20b). Simply substitute into the rule the expressions you have for f(x), g(x), f'(x), and g'(x). You should get:

   u'(x)  =  (m1 (m2x + b2) ) + (m2 (m1 + b1) )
Ordinarily, I wouldn't bother to multiply this out, but I'd like to take this opportunity to demonstrate a method of cross-check. Multiplying the above expression out yields:
   u'(x)  =  m1m2x + m1b2 + m1m2x + m2b1

          =  2m1m2x + m1b2 + m2b1
If we multiply out the original u(x), we get:
   u(x)  =  m1m2x2 + m1b2x + m2b1x + b1b2
Try using the sum rule, the rule about xn, and the rule that about multiplying by a constant on this version of u(x) to demonstrate that the derivative obtained this way is identical to the one we obtained the other way.

4b) Here you are given the product of two quadratics. We do this the same way as we did 4a. Let  f(x) = x2 + 2x + 1  and let  g(x) = x2 - 3x + 2. We can find the derivative of each by applying the xn rule and the sum rule. Hence  f'(x) = 2x + 2  and  g'(x) = 2x - 3. Now simply apply the product rule (equation 4.2-20b), substituting in these expressions for f(x), g(x), f'(x), and g'(x). Doing that yields:

   u'(x)  =  ( (2x + 2) (x2 - 3x + 2) ) + ( (2x - 3) (x2 + 2x + 1) )
which is the answer. If you want to multiply out u'(x) and u(x) to do the cross check, by all means do so. It's good exercise.

As further exercise, observe that the f(x) and g(x) we used in this problem are both, themselves, products, if you factor them:

   f(x)  =  x2 + 2x + 1  =  (x + 1) (x + 1)

   g(x)  =  x2 - 3x + 2  =  (x - 1) (x - 2)
Try applying the product rule to both of those and make sure you come up with the same expressions for f'(x) and g'(x) as we did in the first part of this problem (ie 4b).

4c) This is still the same problem as the two previous ones (ie 4a and 4b). Again, we find that u(x) is a product, and we set f(x) and g(x) to the two factors respectively. So  f(x) = x - 1  and  g(x) = 4x4 - 7x3 + 2x2 - 5x + 8. We find that  f'(x) = 1, which ought to be pretty easy for you by now. To take the derivative of g(x) you have to see it as the sum of a bunch of xn terms, each multiplied by a constant. So apply the xn rule, the rule for multiplying by a constant, and the sum rule, and you get  g'(x) = 16x3 - 21x2 + 4x - 5. When you apply the product rule (equation 4.2-20b) to the f(x), g(x), f'(x), and g'(x) we get here, you find that the answer is:

   u'(x) = ( (1) (4x4 - 7x3 + 2x2 - 5x + 8) ) +

           ( (x - 1) (16x3 - 21x2 + 4x - 5) )

4d) In this one you are given the product of two functions, one, x2, explicitly, the other, f(x), as simply a symbol for any function. So let  g(x) = x2. Using material we have already covered, we can determine that  g'(x) = 2x. So now we apply the product rule (equation 4.2-20b), substituting our expressions for g(x) and g'(x). We can't substitute f(x) or f'(x) because the problem doesn't give anything to substitute. We get as an answer:

   u'(x)  =  (2x f(x) ) + (x2 f'(x) )

4e) This problem is the difference of two products. You have to apply the product rule (equation 4.2-20b) to each product individually to find the derivative of each product. We have already seen that the derivative of the difference is the same as the difference of the derivatives. So to get the answer, simply take the difference of the two derivatives that you got using the product rule. I won't work the details for you here. You should be getting better at this by now. The answer is:

   u'(x)  =  f(x) + xf'(x) - 3x2g(x) - x3g'(x)

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