\[f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]
Use limit to calculate the derivative of a given function. See day10.
Recall that a line can be determined by specifying its slope, say \(k\), and a point \((x_0,y_0)\) on it.
The point-slope form is: \(y-y_0=k(x-x_0)\).
For a given funciton \(f\),
the slope of the tangent line at \(x=x_0\)
is equal to \(f'(x_0)\).
The point \((x_0,f(x_0))\) is on the tangent line.
Hence the equation of the tangent line at \(x_0\) is
\[y-f(x_0)=f'(x_0)(x-x_0),\]
or equivalently,
\[y=f'(x_0)(x-x_0)+f(x_0).\]
Ex. Find the tangent line of \(f(x)=(x-1)^2\) at \(x=2\).
Key: \(y=2(x-2)+1=2x-3\).
(When calculating derivative using limit definition, it's much easier to use \(a^2-b^2=(a+b)(a-b)\) instead of expanding the square terms)
Tips:
Example.
Consider the graph of \(f\) given below. What does the graph of \(f'\) look like?
Starting from the left, \(f\) first decreases, then increases, then decreases again.
So we deduce that \(f'\) is first negative,
then positive, in the end negative again.
The zeros of \(f'\) are where \(f\) has a horizontal tangent line.
See below for a sketch of \(f'\).
See day11.
Cases. \[\frac{0}{0},\quad \frac{\infty}{\infty},\quad\text{others}\]
See Quiz 3.
\(f(x)\) has a horizontal asymptote \(y=y_0\) if \(\lim\limits_{x\to \infty}f(x)=y_0\) or \(\lim\limits_{x\to -\infty}f(x)=y_0\)
For horizontal asymptote, be sure to check both \(x\to \infty\) and \(x\to -\infty\) to find all possible asymptotes.
Examples.
(a). \(f(x)=\frac{1}{x}+9\) has one one horizontal asymptote \(y=9\)
(b). \(f(x)=\frac{3x^2-1}{x^2+2x}\) has one horizontal asymptote \(y=3\)
as \(\lim\limits_{x\to \infty}f(x)=\lim\limits_{x\to -\infty}f(x)=3\)
See day8: http://www.math.emory.edu/~dcai7/teaching.html
Domain of \(f(x)=\ln x\) is \((0,\infty)\), range is \((-\infty,\infty)\) \[\ln \left(e^s\right) = s\] \[\ln a +\ln b = \ln (ab)\] \[\ln a -\ln b = \ln (\frac{a}{b})\]
For more info about \(\exp, \log\), see Quiz 2, day3, day5: http://www.math.emory.edu/~dcai7/teaching.html
See Quiz 1, Quiz 2