Derivative rules [3.1]
Quick
What is the formal definition of the derivative in terms of a limit?
Limit of the Difference Quotient (ldq)
One way we defined the derivative: the limit of the difference quotient (ldq): $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ This should remind you of how we've approximated the derivative numerically. When $h$ is small: $$f'(x)\approx \frac{f(x+h)-f(x)}{h}$$ I Remember this in terms of a line connecting the point $(x, f(x))$ and $(x+h, f(x+h))$. The slope of that line is: $$m=\frac{\Delta f}{\Delta x} =\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h}.$$ Here is a Desmos graph that calculates that approximation for *any* function you type in as $f(x)=...$. For example the function below is $\color{blue}f(x)=x^2.$ Then desmos calculates an aproximation using the $\color{red}\text{ldq with } h=0.02$.
Other interpretations
This is one of several ways of thinking about the derivative. But also...
- Verbal: "The instantaneous rate of change of the function".
- Graphical: The slope of a line that's tangent to the graph (at $(x, f(x)$).
Procedure for today
In what follows we'll take the approach of using the ldq to find the derivative symbolically:
- Write down: "if $f(x)=$... then $f'(x)=$ [ldq for this function]"
- Anyway you can, evaluate the ldq.
- Write your conclusion as: $$\frac{d}{dx}\ \text{[your function]}=\text{...}$$
- We'll use Mathematica for some of these - 3.1.nb
- We may use CoCalc or Desmos to do some graphing to make a guess at some of these.
Constant rule: $f(x)=k$
$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=$$
$$\begineq =&\lim_{h\to 0} \frac{k-k}{h}\\ =&\lim_{h\to 0} \frac{0}{h}=0 \endeq$$ Of course, the fraction $\frac{0}{0}$ is undefined. But when working with limits we don't care what happens right *at* $h=0$. We only need to figure out what's happening as we get closer and closer to 0. So, for finite values of $h$, $0/h$ is always 0, so we could re-write that chain of limits like this..
$$\begineq =&\lim_{h\to 0} \frac{k-k}{h} = \lim_{h\to 0} \frac{0}{h}\\ =&\lim_{h\to 0} 0=\color{blue}0 \endeq$$
| $h$ | 0.1 | 0.01 | 0.001 | 0.0000000000001 |
|---|---|---|---|---|
| $\frac{0}{h}$ | 0 | 0 | 0 | 0 |
Below: $\color{blue}f(x)=3$
Constant rule: $$ \frac{d}{dx}k = 0$$
Identity rule: $f(x)=x$
$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=$$
$$\begineq =&\lim_{h\to 0} \frac{(x+h)-x}{h}=\lim_{h\to 0} \frac{x+h-x}{h}\\ =&\lim_{h\to 0} \frac{h}{h}=1 \endeq$$
Identity rule: $$ \frac{d}{dx}x = 1$$
Quadratic rule:$f(x)=x^2$
$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=$$
$$ \frac{d}{dx}x^2 = 2x$$
Power rule: $f(x)=x^n$
$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=$$
Consult this table:
We showed in class that this implies...
Power rule: $$ \frac{d}{dx}x^n = nx^{(n-1)}$$
Example: $f(x)=\sqrt x$. What is $f'(x)$?
$\sqrt x$ is the same as $x^{1/2}$, so with $n=1/2$ we get:
$$f'(x)=\frac 12 x^{(1/2 -1)}=\frac 12 x^{(-1/2)}=\frac 12 \frac{1}{x^{1/2}}=\color{red}\frac{1}{2\sqrt x}.$$
[Even though we came up with a rule based on $n$ being a positive integer... We'll use the rule you get even for $n$ being negative, or for non-integer values! The proof for this will happen in Math 212.]
Constant Multiple rule: $g(x)=c\,f(x)$
$$g'(x)=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h}=$$
Constant multiple: $$\frac{d}{dx}\left(c\cdot f(x)\right)=c\cdot f'(x).$$
Sum rule: $s(x)=f(x)+ g(x)$
$$s'(x)=\lim_{h\to 0} \frac{s(x+h)-s(x)}{h}=$$
$$\begineq s'(x) =&\lim_{h\to 0} \frac{f(x+h)+g(x+h)-(f(x)+g(x))}{h}\\ =&\lim_{h\to 0} \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\\ =&\lim_{h\to 0} \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h}\\ =&\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}+\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ =&f'(x)+g'(x). \endeq $$
I won't show the difference rule for $d(x)=f(x)-g(x)$. But I think after the sum rule, you could show easily enough that $d'(x)=f'(x)=g(x).$$
Putting these together:
Sum and difference rules: $$\frac{d}{dx}(f(x)\pm g(x))=f'(x)\pm g(x).$$
Example
$f(x)=x^5+3x^4-2x+7$. What is $f'(x)$?
Watch this screencast of taking the...
Derivative of a polynomial (4 minutes)
Starting from $$\frac{d}{dx} \left( x^5+3x^4-2x+7 \right)=$$ First we apply the sum and difference rules: $$= \frac{d}{dx}(x^5)+\frac{d}{dx}(3x^4)-\frac{d}{dx}(2x) + \frac{d}{dx}(7).$$ The last term, $\frac{d}{dx} 7$ is the derivative of a constant, and we know that's 0.
Next we can apply the constant multiple rule: $$= \frac{d}{dx}(x^5)+3\frac{d}{dx}(x^4)-2\frac{d}{dx}(x) .$$
Finally, we will apply the power rule, and simplify the terms: $$\begineq=&5x^4+3*4x^3-2*1\\ =&5x^4+12x^3-2 \endeq$$
Exponential rule: $f(x)=a^x$
$$\begineq f'(x)=&\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac{a^{(x+h)}-a^x}{h}\\ =&\lim_{h\to 0} \frac{a^xa^h-a^x}{h} =\lim_{h\to 0} a^x\frac{a^h-1}{h}\\ =&a^x\lim_{h\to 0}\frac{a^h-1}{h}\\ \endeq $$ You'll complete the derivation of the exponential rule by graphically finding the limit of $\frac{a^h-1}{h}$ in the lab.
Exponential rule: $$\frac{d}{dx}a^x=a^x\cdot \ln a$$
Intuitively: If the relative growth rate is 100% (=1.00), for continuous compounding, doesn't that mean that the growth rate is equal to the amount on deposit?
Homework tip
On the questions that ask you to find a "line tangent to the curve" at a point, since you'll know:- the coordinates of one point that the line needs to pass through (if you know $x$, you can calculate the $y$ coordinate as $f(x)$.)
- the slope, $m$, of that tangent line, which is $f'(x)$, which you can calculate,