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anonymous
 one year ago
Hello, everyone! I would very happy if you helped me.
I can't understand exercise 1C2 in E Readings at all.
Why (f(x)  f(a))/(xa)?
anonymous
 one year ago
Hello, everyone! I would very happy if you helped me. I can't understand exercise 1C2 in E Readings at all. Why (f(x)  f(a))/(xa)?

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phi
 one year ago
Best ResponseYou've already chosen the best response.2they 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 \[ f'(a) = \lim_{x\rightarrow a}\frac{f(x)f(a)}{(xa)} \] both definitions are "change in y" divided by "change in x"

phi
 one year ago
Best ResponseYou've already chosen the best response.2For 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= ua 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 \[ f'(a) = \lim_{u\rightarrow a}\frac{f(u)f(a)}{ua} \] or, renaming u to x (we usually use x as the variable) \[ f'(a) = \lim_{x\rightarrow a}\frac{f(x)f(a)}{xa} \] which is the definition of the derivative evaluated at a specific value x=a

phi
 one year ago
Best ResponseYou've already chosen the best response.2Here is how we get the answer, using the general definition \[ f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)f(x)}{h} \\ f(x)= (xa)g(x) \] plugging in: \[ f'(x) = \lim_{h\rightarrow 0}\frac{(x+ha)g(x+h) (xa)g(x) }{h} \\ =\lim_{h\rightarrow 0}\frac{(xa)g(x+h)+hg(x+h) (xa)g(x) }{h} \\ \lim_{h\rightarrow 0}\frac{(xa)\left(g(x+h)g(x)\right) }{h} +\frac{hg(x+h) }{h} \] the limit of a sum is the sum of limits \[ \lim_{h\rightarrow 0}\frac{(xa)\left(g(x+h)g(x)\right) }{h}+\lim_{h\rightarrow 0}g(x+h) \] in the first term, the limit of a product is the product of the limits \[ \lim_{h\rightarrow 0}(xa)\cdot \lim_{h\rightarrow 0}\frac{g(x+h)g(x)}{h}+\lim_{h\rightarrow 0}g(x+h) \] taking the limits, we get \[ f'(x)= (xa) g'(x) + g(x) \] evaluate at x=a to get \[ f'(a)= g(a) \] using the more specific definition gives a shorter derivation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much! Now everything is clear

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where did the g come from?

phi
 one year ago
Best ResponseYou've already chosen the best response.2@sylviawest this is problem 1C2 in http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/readings/e_exrcs_scsn_1_7.pdf
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