Question

Problem set 1: 1C-2 Let f(x) = (x − a)g(x). Use the definition of the derivative to calculate that
f�
(a) = g(a), assuming that g is continuous.

The solution does not make sense...anyone know?

Answer 1

This is extremely confusing because the problem expects you to know a different form for the definition of the derivative than the one used in this course, probably because this problem set was originally intended to be used in a different calculus course. In this course, we're used to working with this version:\[\lim_{\Delta x \rightarrow 0}\frac{ f(x+\Delta x)-f(x) }{ \Delta x }\]For purposes of this exercise, we want to use this version:\[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]It may not be immediately obvious that these two expressions are equivalent, but if you think about it you'll probably see the relationship. In the first one, the change in x is shown as Delta x, and in the second, it's x-a. As x approaches a, this change in x approaches zero, so that's the same as saying Delta x approaches zero.

Now, when we plug the function given in this exercise into this version of the definition of the derivative, we get\[\lim_{x \rightarrow a}\frac{ (x-a)g(x)-(a-a)g(a) }{ x-a }\]The right half of the numerator is zero because it has zero as a factor (that is, a minus a). Then we're left with\[\lim_{x \rightarrow a}\frac{ (x-a)g(x) }{ x-a }\]We can divide top and bottom by (x-a) because we're interested in the limit as x approaches a, and not the value where x=a (where we would be dividing by zero). That leaves us with just g(x), and the limit as x approaches a is g(a), and that's what you're asked to show.

Answer 2

Thank you so much! I was entirely confused because I didn't remember seeing the definition for that style of the derivative anywhere else.

Answer 3

The problem defines f(x) and then asks for f'(a).
How do you get the function f(a) as opposed to f(x)?

Answer 4

We're speaking loosely when we say f(x) is a function. Strictly speaking, f is a function, and f(x) is what we get when we apply this function to f. So f(a) is the same function, but applied to a now instead of x. In this problem we're being asked to use the definition of the derivative (in the format I described above) to find the derivative of this function, evaluated at a.

Perhaps the more interesting question is why we get this result. This boils down to application of the product rule. You have a product of (x-a) and some function. Applying the product rule to find the derivative, you get 1 times the function (because the derivative of (x-a) is 1) plus the derivative of the function times zero (because (x-a) equals zero when evaluated at a).

Answer 5

Ah, thanks. This one was tricky due to the notation.