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Homework Help: Simple derivative question

  1. Dec 4, 2007 #1
    hi. i need some help with a derivative question. i can get the answer and all but it takes a long time to do it.

    i need to find the instantaneous rate of change expression (derivative), and i MUST use the first principle.


    i can do this with the chain rule method quite easily. the first principle method takes forever as i use pascal's triangle to expand and the brackets and all.

    is there some sort of a quicker method where by substituting in some other variable or something. i really don't want to do ten or so of these that take 10 minutes to get through.
  2. jcsd
  3. Dec 4, 2007 #2
    Can you clarify (i.e. state) the first principle? This term is unfamiliar to some of us without further context. Do you mean using the limit definition of the derivative?
  4. Dec 4, 2007 #3

    lim h[tex]\rightarrow[/tex]0
  5. Dec 4, 2007 #4
    I'll start off by saying: what an awful assignment. One can generally assess whether or not a student understands the limit definition without resorting to such tedious assignments in symbolic manipulation.

    That said, you are probably doing it correctly; and yes, it is tedious. However, you may be able try the following (via the limit definition):

    The chain rule can be derived from the limit definition as follows:

    [tex]\lim_{h\to 0} \frac{f[g(x+h)] - f[g(x)]}{(x+h)-x}
    = \lim_{h\to 0} \left( \frac{f[g(x+h)] - f[g(x)]}{g(x+h)-g(x)} \cdot \frac{g(x+h)-g(x)}{(x+h)-x}\right)[/tex]

    Using the properties of products of limits, we obtain:

    [tex] \left( \lim_{h\to 0} \frac{f[g(x+h)] - f[g(x)]}{g(x+h)-g(x)} \right)
    \cdot \left(\lim_{h\to 0} \frac{g(x+h)-g(x)}{(x+h)-x}\right) = f'[g(x)]\cdot g'(x)[/tex]

    In this last equation, the [tex]g(x+h)[/tex] and [tex]g(x)[/tex] are now the "x-coordinates" and the [tex]f[g(x+h)][/tex] and [tex]f[g(x)][/tex] are the corresponding "y-coordinates" (notice that the fraction is essentially the slope through the two "points"). Perhaps you may be allowed to taylor this derivation to your own problems in order to reduce the amount of symbolic manipulation.
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