Limits, Continuity, and the Definition of the Derivative [2.FT]
Test 1 summary
I changed the grades on Test 1 in moodle to be out of 103 (instead of 105) possible points.
The average was 81%.
- 90% < 6 people
- 80% < 3 < 90%
- 70% < 5 <80%
- 00% < 3 < 70%
Chapter 2 Focus on Theory (pages 127-132).
Rational Functions (lab notes) (.nb)
For today's class, you'll likely be using a calculator: Make sure that your calculator is set to Radians, not Degrees.
-OR-
Use CoCalc as your 'calculator' for today. (Set to Radians by default)
[In this class, you should assume that the input, $x$, is always in radians for trig functions like $\sin x$, $\cos x$, $\tan x$, unless degrees are explicitly asked for.]
Handout (Mathematica .nb file);
- Handout (.pdf file),
- Starship Enterprise graph (.pdf),
- Handout answers (.pdf)
Analytically, the derivative of the function $f(x)$ is the derivative function, $f'(x)$ which is $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
Example
How to use that analytic definition to calculate the derivative of $f(x)=x^2 -4x +7$?:
Numerical approximations for the derivative:
It looks like the values are getting closer and closer to 2 (or "approaching a limit of" 2), as $h$ gets smaller and smaller.
Symbolically, using the definition of the derivative...
- Since $f(x)=x^2-4x+7$,
$f(x+h)=(\ x+h\ )^2-4*(\ x+h\ )+7=x^2+2xh+h^2 -4x-4h +7$. - So, $f(x+h)-f(x)$ can be calculated as: $$\begineq f(x+h) -[f(x)]=&x^2+2xh+h^2-4x-4h+7-[x^2-4x+7]\\ =&\color{blue}{x^2}\color{black}+2xh+h^2\color{red}{-4x}\color{black}-4h\color{orange}{+7} \color{blue}{-x^2}\color{red}{+4x}\color{orange}{-7}\\ =&2xh-4h+h^2\\ \endeq $$ The terms with the same colors cancel out.
- Next, calculate the difference quotient: $$\begineq \frac{f(x+h)-f(x)}{h}=&\frac{2xh-4h+h^2}{h}\\ =&\frac{2xh}h-\frac{4h}h+\frac{h^2}h\\ =&2x-4+h\\ \endeq $$
- We are ready to take the limit of the difference quotient, as $h\to 0$: $$ \begineq \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=& \lim_{h\to 0}\left[2x-4+h\right]\\ =&\lim_{h\to 0} 2x-\lim_{h\to 0}4+\lim_{h\to 0}h=2x-4+0\\ f'(x)=&2x-4 \\ \endeq $$