Question

Hello, everyone! I would very happy if you helped me.
I can't understand exercise 1C-2 in E Readings at all.
Why (f(x) - f(a))/(x-a)?

Answer 1

they are giving the "short version"
we could use this definition of the derivative
\[ f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} \]
In the above, the denominator is a simplification of (x+h)-x = h

or, if we are only interested in f'(a), we can use this more specific definition
\[ f'(a) = \lim_{x\rightarrow a}\frac{f(x)-f(a)}{(x-a)} \]

both definitions are "change in y" divided by "change in x"

Answer 2

For what it is worth, start with the general definition
\[ f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} \]
let x=a (a fixed value),
and u= x+h= a+h ( to avoid confusion, we will use a different variable name u here)
also, from u=a+h, we get h= u-a

next we note that as h ->0 then u= a+h approaches a
thus we can replace the limit as h->0 with the limit as u->a

substituting in we get
\[ f'(a) = \lim_{u\rightarrow a}\frac{f(u)-f(a)}{u-a} \]
or, renaming u to x (we usually use x as the variable)
\[ f'(a) = \lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} \]
which is the definition of the derivative evaluated at a specific value x=a

Answer 3

Here is how we get the answer, using the general definition
\[ f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} \\ f(x)= (x-a)g(x) \]
plugging in:
\[ f'(x) = \lim_{h\rightarrow 0}\frac{(x+h-a)g(x+h) -(x-a)g(x) }{h} \\
=\lim_{h\rightarrow 0}\frac{(x-a)g(x+h)+hg(x+h) -(x-a)g(x) }{h} \\
\lim_{h\rightarrow 0}\frac{(x-a)\left(g(x+h)-g(x)\right) }{h} +\frac{hg(x+h) }{h}
\]
the limit of a sum is the sum of limits
\[ \lim_{h\rightarrow 0}\frac{(x-a)\left(g(x+h)-g(x)\right) }{h}+\lim_{h\rightarrow 0}g(x+h) \]
in the first term, the limit of a product is the product of the limits
\[ \lim_{h\rightarrow 0}(x-a)\cdot \lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}+\lim_{h\rightarrow 0}g(x+h) \]
taking the limits, we get
\[ f'(x)= (x-a) g'(x) + g(x) \]
evaluate at x=a to get
\[ f'(a)= g(a) \]

using the more specific definition gives a shorter derivation.

Answer 4

Thank you very much! Now everything is clear

Answer 5

Where did the g come from?

Answer 6

@sylviawest
this is problem 1C-2 in
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/readings/e_exrcs_scsn_1_7.pdf