Derivative from first principles

A derivative measures the rate of change of a function. The formal definition is a limit: $\boxed{f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}.}$

This is called the derivative from first principles (or "ab initio").

Geometric meaning

The quantity $\dfrac{f(a + h) - f(a)}{h}$ is the slope of the secant line through $(a, f(a))$ and $(a + h, f(a + h))$. As $h \to 0$, the secant rotates and (in the limit) becomes the tangent line at $(a, f(a))$.

So $f'(a)$ = slope of the tangent line to $y = f(x)$ at $x = a$.

Physical meaning

If $s(t)$ is the position of a particle at time $t$, then $\dfrac{s(t + h) - s(t)}{h}$ is the average velocity over the time interval $[t, t + h]$. As $h \to 0$, this becomes the instantaneous velocity at time $t$: $v(t) = s'(t).$

Notation

• $f'(a)$ , Lagrange notation.
• $\dfrac{df}{dx}\bigg|_{x = a}$ or $\dfrac{dy}{dx}\bigg|_{x = a}$ , Leibniz notation.
• $\dot{f}(a)$ , Newton's dot notation, common in physics.

When $f'$ is well-defined for every $a$ in some interval, we have a new function $f'$, called the derivative of $f$.

Computing from first principles

To find $f'(x)$ for a given $f$:

1. Write $\dfrac{f(x + h) - f(x)}{h}$.
2. Simplify (algebra, factor, etc.).
3. Take the limit $h \to 0$.

Worked examples

Example 1. Find $f'(x)$ for $f(x) = x^2$.

$\dfrac{f(x + h) - f(x)}{h} = \dfrac{(x + h)^2 - x^2}{h} = \dfrac{2 x h + h^2}{h} = 2 x + h \to 2 x$ as $h \to 0$.

So $(x^2)' = 2 x$.

Example 2. Find $f'(x)$ for $f(x) = x^3$.

$\dfrac{(x + h)^3 - x^3}{h} = \dfrac{3 x^2 h + 3 x h^2 + h^3}{h} = 3 x^2 + 3 x h + h^2 \to 3 x^2$.

So $(x^3)' = 3 x^2$. Pattern: $(x^n)' = n x^{n-1}$.

Example 3. Find $f'(x)$ for $f(x) = \sqrt{x}$.

$\dfrac{\sqrt{x + h} - \sqrt{x}}{h} = \dfrac{(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})} = \dfrac{h}{h(\sqrt{x + h} + \sqrt{x})} = \dfrac{1}{\sqrt{x + h} + \sqrt{x}} \to \dfrac{1}{2 \sqrt{x}}$.

So $(\sqrt{x})' = \dfrac{1}{2 \sqrt{x}}$.

Example 4. Find $f'(x)$ for $f(x) = \sin x$.

$\dfrac{\sin(x + h) - \sin x}{h} = \dfrac{\sin x \cos h + \cos x \sin h - \sin x}{h} = \sin x \cdot \dfrac{\cos h - 1}{h} + \cos x \cdot \dfrac{\sin h}{h}$.

As $h \to 0$: $\dfrac{\cos h - 1}{h} \to 0$ and $\dfrac{\sin h}{h} \to 1$. So the limit is $\sin x \cdot 0 + \cos x \cdot 1 = \cos x$.

Thus $(\sin x)' = \cos x$.

Example 5. Find $f'(x)$ for $f(x) = 1/x$.

$\dfrac{1/(x + h) - 1/x}{h} = \dfrac{x - (x + h)}{h x (x + h)} = \dfrac{-h}{h x (x + h)} = \dfrac{-1}{x(x + h)} \to \dfrac{-1}{x^2}$.

So $(1/x)' = -1/x^2$.

Try it yourself

Find $f'(x)$ from first principles:

1. $f(x) = 3 x + 5$.
2. $f(x) = x^2 + 1$.
3. $f(x) = x^2 - 3 x$.
4. $f(x) = 5$ (constant).
5. $f(x) = 1/x^2$.
6. $f(x) = \cos x$. (Hint: similar to $\sin x$.)
7. $f(x) = x^4$.
8. $f(x) = (x + 1)^2$.
9. $f(x) = \sqrt{x + 1}$.
10. $f(x) = x + 1/x$. (For $x \ne 0$.)
11. $f(x) = \tan x$ at $x = 0$.
12. $f(x) = 1/\sqrt{x}$ at $x = 4$.

Pitfalls / Tricks

• The "$h \to 0$" is essential , the difference quotient is undefined at $h = 0$.
• Always simplify before taking the limit (so the $h$ in the denominator can be cancelled).
• For trig derivatives, use $\sin(A + B) = \sin A \cos B + \cos A \sin B$ to expand.
• Insight. The derivative is the linear approximation: near $x = a$, $f(x) \approx f(a) + f'(a)(x - a)$. This is the tangent line equation.