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anonymous

  • one year ago

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)?

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  1. phi
    • one year ago
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    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"

  2. phi
    • one year ago
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    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

  3. phi
    • one year ago
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    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.

  4. anonymous
    • one year ago
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    Thank you very much! Now everything is clear

  5. anonymous
    • one year ago
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    Where did the g come from?

  6. phi
    • one year ago
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    @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

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