Find $f'(a)$ if $f(x) = \frac {x^n - a^n}{x-a}$

Question

I'll state the multiple choice question from my textbook below:

If $f(x) = \frac {x^n - a^n}{x - a}$ for some constant '$a$', then $f'(a)$ is

(A) $1$

(B) $0$

(C) does not exist

(D) $\frac 12$

Here's what I tried:

By using quotient rule we get:

$f'(x) = \frac {(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2}$

Clearly,

$f'(a) = \frac 00$

And the correct choice according to my book is (C). Does getting $\frac 00$ for a particular value of the variable (in this case, for $x=a$) mean that the derivative of the function doesn't exist at that point?

Not satisfied by the answer I factorised the numerator of $f(x)$ and cancelled the factor $(x-a)$ as below:

$f(x) = \frac {(x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 +..........+ xa^{n-2} + a^{n-1})}{x-a}$

$\implies f(x) = x^{n-1} + x^{n-2}a + x^{n-3}a^2 +..........+ xa^{n-2} + a^{n-1}$

Differentiating with respect to $x$ we get,

$f'(x) = (n-1)x^{n-2} + (n-2)x^{n-3}a + (n-3)x^{n-4}a^2 +..........+ a^{n-2}$

$\implies f'(a) = \Big[(n-1) + (n-2) + (n-3) + .......... + 1\Big]a^{n-2}$

$\implies f'(a) = \frac {n(n-1)}2 a^{n-2}$

Now this is a completely different answer.

So what have I done wrong? Is there some problem with cancelling the factor $(x-a)$? Does it have something to with the continuity and differentiabilty of $f(x)$ at $a$? What if the function was $f(x) = \begin{cases} \frac {x^n - a^n}{x - a}, & \text{if $x \ne a$} \\ na^{n-1}, & \text{if $x = a$} \end{cases}$? How do I find the derivative at $x = a$ in this case?

Answer 1

For the first case the derivative $f'(a)$ doesn't exist since $f(x)$ is not defined for $x=a$ and the derivative is defined by

$$\lim_{x\to a} \frac{f(x)-\color{red}{f(a)}}{x-a}$$

and the existence of $f(a)$ is required.

In the second case we are allowed to apply the definition by limit.

Answer 2

The second way is correct. The problem with the first way is that the definition of $f$ doesn't hold at $a$. Presumably, the text means for the definition of $f$ to be extended to $a$ by continuity. Then you would take the limit of the difference quotient as $x\to a.$ When you get $\frac{0}0{}$ as a limit, you can't conclude that the limit doesn't exist. You'll learn about these "indeterminate forms" shortly, if you haven't already.

Answer 3

If the correct answer is "C" then they are being very technical. The function $f$, as written, is not defined at $x = a$. So there's no hope of evaluating the limit $\dfrac{f(x) - f(a)}{x-a}$ — you don't have a value to put in for $f(a)$.

You have realized that the not-being-defined discontinuity at $x = a$ is removable, and in the second half of your question you've "rewritten" $f$ in a way that removes the discontinuity. I think that gives the most insight into what's happening, but technically your re-written $f$ is a different function than the original $f$ — one of them is defined at $a$ and one of them isn't.

Our human minds want to tell us to "redefine $f$ at any removable discontinuity so that it is continuous there", but rigorously mathematically it doesn't happen unless it is explicitly stated. It is not explicitly stated in the textbook's question, so "C" is the technically correct answer.

Now, as Ennar points out, it is important that you learn that the domain is an essential part of the definition of a function, especially when you are first starting to work with functions in calculus. Also, not explicitly giving the domain of a function happens a lot in math, so you have to get good at dealing with situations like this, and in a situation like this the technical answer is the correct answer.

As you go on and develop more tools and better math language you should deal less with multiple choice questions and more with ones that ask "Describe what happens if you want to differentiate the function $\dfrac{x^n-a^n}{x-a}$ at $x=a$.", and then it will be appropriate to do what you did in the second half of your question.