Calculus Unit 3: The Derivative | Notes and practice

Notes

Average rate of change from $a$ to $b$ is the secant slope $\displaystyle \frac{f(b) - f(a)}{b - a}$.
The instantaneous rate of change is the limit as the interval shrinks. It is the slope of the tangent line.

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.
Known results: $(x^2)' = 2x$, $(x^3)' = 3x^2$, $\displaystyle \left(\frac{1}{x}\right)' = -\frac{1}{x^2}$, $\displaystyle (\sqrt{x})' = \frac{1}{2\sqrt{x}}$.
Positive $f'$ means increasing, negative means decreasing, zero means a horizontal tangent.

$f'(x)$, $y'$, $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d}{dx}[f(x)]$ all mean the derivative.
Corners, cusps, vertical tangents and discontinuities make a function non-differentiable. Differentiable implies continuous, but not the other way around.
Tangent line at $x = a$: $y - f(a) = f'(a)(x - a)$.

Each topic has its own short notes and practice.

Find the average rate of change of $f(x) = x^3$ from $x = 1$ to $x = 2$.

Use the limit definition to find $f'(x)$ for $f(x) = 2x^2$.

What is the slope of the tangent line to $\displaystyle f(x) = \sqrt{x}$ at $x = 9$? (Use $\displaystyle f'(x) = \frac{1}{2\sqrt{x}}$.)

At which point is $f(x) = |x - 2|$ not differentiable?

A function has $f(3) = 10$ and $f'(3) = -2$. What is the tangent line at $x = 3$?

If $s(t)$ is the position of a runner at time $t$, what does $s'(t)$ represent?

Write your answer first, then open the scoring guide.

Free-response question

Let $f(x) = x^2 + 3x$.

(a) Find the average rate of change of $f$ from $x = 1$ to $x = 3$.

Show the answer and scoring guide

$f(1) = 4$ and $f(3) = 18$. The average rate is $\displaystyle \frac{18 - 4}{2} = 7$.

Check off the points you earned:

(b) Use the limit definition to find $f'(x)$.

Show the answer and scoring guide

$\displaystyle \frac{f(x + h) - f(x)}{h} = \frac{2xh + h^2 + 3h}{h} = 2x + h + 3$. As $h \to 0$, $f'(x) = 2x + 3$.

Check off the points you earned:

(c) Write the equation of the tangent line at $x = 1$.

Show the answer and scoring guide

$f'(1) = 5$ and $f(1) = 4$, so $y - 4 = 5(x - 1)$, or $y = 5x - 1$.

Check off the points you earned: