**1)** In the first expression we have the sum of two functions,
each of which
can be expressed
as `x ^{n}`, one where

In the second expression we have the sum of an `x ^{n}` and
an

**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``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

g(x) = (1) f(x) = f(x)hence

g'(x) = f'(x) = (1) f'(x)So it works for

gwhenever_{n}'(x) = n f'(x)

gthen_{n}(x) = n f(x)

gThat the derivative of_{n+1}(x) = (n + 1) f(x) = (n f(x) ) + f(x)

gAnd you have it proved by induction. But you only need to do it that way if you are performing for a stickler on formality._{n+1}'(x) = (n f'(x) ) + f'(x) = (n + 1) f'(x)

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

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

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)`` u(x) + g(x) = f(x)`

email me at *hahn@netsrq.com*

**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 `` f(x) = m _{1}x + b_{1}`

u'(x) = (mOrdinarily, 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:_{1}(m_{2}x + b_{2}) ) + (m_{2}(m_{1}+ b_{1}) )

u'(x) = mIf we multiply out the original_{1}m_{2}x + m_{1}b_{2}+ m_{1}m_{2}x + m_{2}b_{1}= 2m_{1}m_{2}x + m_{1}b_{2}+ m_{2}b_{1}

u(x) = mTry using the sum rule, the rule about_{1}m_{2}x^{2}+ m_{1}b_{2}x + m_{2}b_{1}x + b_{1}b_{2}

**4b)** Here you are given the product of two quadratics. We do this
the same way as we did 4a. Let
` f(x) = x ^{2} + 2x + 1 `

u'(x) = ( (2x + 2) (xwhich is the answer. If you want to multiply out^{2}- 3x + 2) ) + ( (2x - 3) (x^{2}+ 2x + 1) )

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) = xTry applying the product rule to both of those and make sure you come up with the same expressions for^{2}+ 2x + 1 = (x + 1) (x + 1) g(x) = x^{2}- 3x + 2 = (x - 1) (x - 2)

**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 `` g(x) = 4x ^{4} - 7x^{3} + 2x^{2} - 5x + 8`

u'(x) = ( (1) (4x^{4}- 7x^{3}+ 2x^{2}- 5x + 8) ) + ( (x - 1) (16x^{3}- 21x^{2}+ 4x - 5) )

**4d)** In this one you are given the product of two functions, one,
`x ^{2}`, explicitly, the other,

u'(x) = (2x f(x) ) + (x^{2}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) - 3x^{2}g(x) - x^{3}g'(x)

email me at *hahn@netsrq.com*