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\centerline{Mathematics 133 -- Intensive Calculus for Science}
\centerline{Discussion 3 -- The Derivative Function, Graphically, Numerically,
and Symbolically}
\centerline{October 4, 2000}
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{\it Background}
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We have seen how the derivative of a function $f(x)$ at $x = a$, 
$$f'(a) = \lim_{h\to 0} {f(a+h) - f(a) \over h}$$
measures the {\it slope of the graph} $y = f(x)$ at $x = a$ (that is,
the slope of the tangent line to $y = f(x)$ at $x = a$).  Letting
$x = a$ vary, we get the {\it derivative function} $f'(x)$:
$$f'(x) = \lim_{h\to 0} {f(x+h) - f(x) \over h}.\leqno(1)$$
Today, we want to look at some additional examples of derivative 
functions graphically, numerically, and symbolically.
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\noindent
{\it Discussion Questions}
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\item{A)}  The following graph shows $y = f(x)$ for some function.
At which $x$ is $f'(x) = 0$?  On which intervals is $f'(x) > 0$?  
On which intervals is $f'(x) < 0$?  Using this information, give 
a rough sketch of the graph $y = f'(x)$.  
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\item{B)}  Suppose we wanted to understand $f'(x)$ for the function
$f(x) = 6^x$ (exponential growth).  Using (1) above and our rules
for exponents, explain why
$$f'(x) = 6^x \cdot \lim_{h\to 0} {6^h - 1\over h}$$
To go farther at this point, we need to resort to some numerical work
using a calculator.  Compute ${6^h - 1\over h}$ for $h = \pm 1, \pm .1, \pm .01,
\pm .001$.  Using this information, guess an approximate value for 
$\lim_{h\to 0} {6^h - 1\over h}$ good to 2 decimal places.  Then this constant
times $6^x$ gives an approximation to the derivative function of $f(x) = 6^x$.
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\item{C)}  What is $f'(x)$ for $f(x) = {1\over x}$?  Use the limit definition:
$$\eqalign{f'(x) &= \lim_{h\to 0} {f(x+h) - f(x)\over h}\cr
                 &= \lim_{h\to 0} {{1\over x+h} - {1\over x}\over h}\cr}$$
then try to simplify and take the limit as $h \to 0$.
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{\it Assignment}
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Group writeups due on Wednesday, October 11 (this is the first day back
from Columbus Day break).
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