Lecture · 18 min · intermediate

The derivative

Instantaneous rate, and the best linear approximation

Courant · the derivative

Average speed on $[t, t + h]$ is $\frac{s ( t + h ) - s ( t )}{h}$ . Instantaneous speed is what that average approaches as the window $h$ shrinks. The same number is the slope of the tangent line — the unique line that hugs the graph better than any other line through the point. The derivative is that number, as a function of the point.

On the board today

—Write the definition of $f^{'} (a)$ as a limit of difference quotients.
—Compute $f^{'}$ for powers from the definition, then use the first rules.
—Read $f^{'} (a)$ as slope and as instantaneous rate.

Definition

f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}

If the limit exists, $f$ is differentiable at $a$ .

Differentiable implies continuous

If $f^{'} (a)$ exists, then $f$ is continuous at $a$ .

f (a + h) - f (a) = f^{'} (a) h + o (h)

The change in $f$ is roughly a constant times $h$ , which goes to $0$ with $h$ . Continuity is cheaper than differentiability: $∣ x ∣$ is continuous at $0$ but not differentiable there.

Worked example · From the definition

Differentiate $f (x) = x^{2}$ at $x = 3$ .

1.Difference quotient: $\frac{( 3 + h ) ^{2} - 9}{h} = \frac{6 h + h ^{2}}{h} = 6 + h$ .
2.Let $h \to 0$ : $f^{'} (3) = 6$ .

More generally $(x^{2})^{'} = 2 x$ . The tangent at $3$ is $y - 9 = 6 (x - 3)$ .

Pause the lecture

Why is $∣ x ∣$ not differentiable at $0$ ?

Common mistake. Writing $\frac{d}{d x} (uv) = u^{'} v^{'}$ . The product rule is $(uv)^{'} = u^{'} v + u v^{'}$ . Forgetting the second term is the calculus version of forgetting $2 ab$ in $(a + b)^{2}$ .

Practice this lecture

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