Computing a derivative at a point one point at a time is slow. If we keep $x$ general instead of picking a number, the same limit produces a formula for the slope at every point. That formula is itself a function.
Key idea
The derivative function is $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ for every $x$ where the limit exists. Its graph shows the slope of $f$: $f' > 0$ where $f$ rises, $f' < 0$ where $f$ falls, and $f' = 0$ where the tangent is flat.
Why it works
The slope of $f$ can be different at different points, so it depends on $x$: that is what makes it a function. The definition is the same limit as before, with $a$ replaced by a letter. Every formula you will learn in the next unit comes from evaluating this limit once, in general, so you never have to do it again for that kind of function.
Worked example: A quadratic, in general
Find $f'(x)$ for $f(x) = 3x^2 - 5x$.
- $$f(x + h) = 3(x + h)^2 - 5(x + h) = 3x^2 + 6xh + 3h^2 - 5x - 5h$$.
- Subtract $f(x) = 3x^2 - 5x$: $6xh + 3h^2 - 5h$.
- Divide by $h$: $6x + 3h - 5$.
- Let $h \to 0$: $f'(x) = 6x - 5$.
Answer: $f'(x) = 6x - 5$. For example, the slope at $x = 1$ is $1$ and at $x = 0$ it is $-5$.
Worked example: A reciprocal, in general
Find $f'(x)$ for $f(x) = \frac{1}{x}$.
- $\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}$.
- Divide by $h$: $\frac{-1}{x(x + h)}$.
- Let $h \to 0$: $-\frac{1}{x^2}$.
Answer: $f'(x) = -\frac{1}{x^2}$, negative everywhere: $\frac{1}{x}$ falls on each side of 0.
Reading $f'$ from the graph of $f$
- Where $f$ climbs, $f'$ is above the axis; where $f$ falls, below.
- At a peak or a valley of a smooth $f$, $f' = 0$.
- Where $f$ is steepest, $|f'|$ is largest.
Common mistake
Sketching $f'$ so that it looks like $f$: a peak in $f$ drawn as a peak in $f'$. A peak in $f$ is where the slope is 0, so $f'$ crosses the axis there. Check each turning point of $f$: it must line up with a zero of $f'$.