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Let's look at #54: it's kind of like the Tibetan monk problem, in that we had to create a single function from the two functions (up and down).
Let's use the definition of the derivative function,
\[ f^\prime(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]
to determine the derivative functions of these important functions:
The first two derivatives are easy: each function has a straight line graph, so the tangent line is the graph of the function itself. The slope of each of those graphs is easy to see (either 0 or m). The derivatives are hence on full display. Not so much for the quadratic, and it's the first function that has an explicit dependence on $x$.
How well can you recognize derivatives and the functions that give rise to them? Let's try #3, p. 122.
Some thoughts:
Let's think of a quadratic motion, e.g. the motion of an eraser thrown across the room. Let $s(t)$ be the height of the eraser at time $t$:
\[s(t)=at^2+bt+c\]
Each of the coefficients has an important, intuitive role to play:
Think about the parabola which is this function's graph: it's an umbrella. The ball will rise, reach its apex, then fall back to Earth.
\[s(t)=40t-16t^2\]
\[s'(t)=v(t)=40-32t\] \[s''(t)=v'(t)=a(t)=-32\] |
Blue -- graph of $s(t)$
Red -- graph of $s'(t)$ |
The parentheses are supposed to tip you off to the fact that these aren't powers, but rather derivatives.
\[ f^\prime(x) = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}\left(f(x)\right)=\left(f(x)\right)'=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]
There are several different ways of writing the derivative, and you need to get used to them. The form \(\frac{d}{dx}\left(f(x)\right)\) reminds us that the operation of differentiation is itself a function: it takes a function in its domain and returns another function -- the derivative function.
To prove it we use one of my favorite tricks: the addition of a special form of 0.
where $n$ is a positive integer.
We can prove this via dominoes -- i.e., mathematical induction -- or via the binomial theorem)
The derivative of the monomial $x^n$ is $nx^{n-1}$, and
The derivative of the monomial $c x^n$ is $nc x^{n-1}$ (by constant multiple).
A polynomial is just a sum of these. So we apply the sum rule, and the power rule, and the constant multiple rule to the flight of the eraser, to get
$s'(t)=(at^2+bt+c)'=2at+b$
and
$s''(t)=(2at+b)'=2a$
$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$