Question

There is a question in the first exercise file:
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 says:
(f(x)-f(a))/(x-a) = ((x-a)g(x)-0)/(x-a) = g(x)->g(a)
as x -> a
I couldn't understand the solution from the very beginning. Where the formula comes from and why its denominator is x-a?
Thanks for any help

Answer 1

can you post a link to the question?

To answer the question, you need to learn the definition of derivative
Have you learned it ?

Answer 2

Yes, I know the definition of derivative but I couldn't relate the formula to this definition. And the main reason for this, that the denominator of formula is x-a.
I couldn't add a link but I attach the exercise and solution pdfs. The question number is 1C-2.
Thanks for your help.

Answer 3

the derivative may be thought of as "change in y" divided by "change in x" i.e. the slope of the line connecting the points (x, f(x)) and (x+h, f(x+h) )
Here the y value is f(x), and h is a small increment (that we let approach 0)
Thus one way to write the definition of derivative is
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x)}{ (x+h) - x} = \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x)}{ h} \]
in other words, the derivative is the slope between two points on the curve that are *very* close to each other. We let h approach (but *never* reach) zero.

I would answer the question using the above definition.
Begin with \( f(x) = (x-a) g(x) \) so
\[ f'(a) = \lim_{h \rightarrow 0} \frac{ f(a+h) - f(a)}{ h} = \lim_{h \rightarrow 0} \frac{ ((a+h)-a) g(a+h) - \cancel{(a-a)}0\ g(a)}{ h} \\
=\lim_{h \rightarrow 0} \frac{ \cancel{h} \ g(a+h) }{ \cancel{h}} \\
=\lim_{h \rightarrow 0} g(a+h) = g(a)
\]

Answer 4

Another way to write the definition of the derivative is to pick two points, say a and x
and let x approach a. This is equivalent to letting x= a+h , and letting h approach 0.
However using this version , the definition is
\[ f'(a) = \lim_{x \rightarrow a} \frac{ f(x) - f(a)}{ x - a} \]
notice this is equivalent to the definition given up above: change in y (i.e. change in f(x)-f(a)) divided by change in x (i.e. x-a)

Answering the question
\[ f'(a) = \lim_{x \rightarrow a} \frac{ (x-a)g(x) - \cancel{(a-a)}0\ g(a) }{ x - a} \\
= \lim_{x \rightarrow a} \frac{ (x-a)g(x) }{ (x - a)}
\]
we let x approach (but *never* reach) a, so x-a is non-zero, and we can divide (x-a)/(x-a)
(notice that if x became a, we would have 0/0, which is indeterminate, and we could not simplify... thus we never allow x to become a)
\[ = \lim_{x \rightarrow a} \frac{ \cancel{(x-a)}g(x) }{ \cancel{(x - a)}} \\
= \lim_{x \rightarrow a} g(x) = g(a)
\]

Answer 5

Thank you very much for your detailed explanation. It helped me a lot.