Modules › The derivative
The derivative
Definition as a limit, tangent line, rates of change, differentiability versus continuity, higher derivatives.
Lecture 3 · Tutorial 3 (second half) · Derivatives
Textbook sections: Thomas ch. 3 Derivatives (3.1–3.11) · Stewart ch. 2 (2.7–2.8) and ch. 3 Differentiation Rules (3.1–3.5, 3.9, 3.10)
Textbook sections: Thomas ch. 3 Derivatives (3.1–3.11) · Stewart ch. 2 (2.7–2.8) and ch. 3 Differentiation Rules (3.1–3.5, 3.9, 3.10)
The derivative
Calculus I · Module 4
After this module you can
- Compute $f'(a)$ from the limit definition and write the tangent line.
- Explain why differentiable implies continuous but not conversely (corners, cusps, vertical tangents).
- Interpret $f'$, $f''$ as velocity and acceleration.
Definition and meaning
- $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$.
- Geometric meaning: slope of the tangent line; tangent line $y=f(a)+f'(a)(x-a)$.
- Physical meaning: instantaneous rate of change; $v=s'$, $a=v'=s''$; speed $=|v|$.
- Example: $f(x)=\sqrt x$ at $4$: $f'(4)=\frac{1}{2\sqrt4}=\frac14$, tangent $y=\frac x4+1$.
When is a function not differentiable?
| Reason | Example at $x=0$ | One-sided derivatives |
|---|---|---|
| corner | $|x|$ | $-1$ and $+1$ |
| vertical tangent | $\sqrt[3]{x}$ | slope $\to+\infty$ |
| cusp | $x^{2/3}$ | $-\infty$ and $+\infty$ |
| discontinuity | $\frac{|x|}{x}$ | not defined |
Theorem
Differentiable at $a$ $\Rightarrow$ continuous at $a$. The converse is false ($|x|$).Squeeze for derivatives
- If $|f(x)|\le x^2$ near $0$ then $f(0)=0$ and $\left|\frac{f(h)}{h}\right|\le|h|\to0$: $f'(0)=0$.
- $f(x)=x^2\sin\frac1x$, $f(0)=0$: $f'(0)=0$ but $f'(x)=2x\sin\frac1x-\cos\frac1x$ has no limit at $0$: $f'$ is not continuous.
- Differentiability of a piecewise function at the junction: continuity AND equal one-sided derivatives.
Rates of change
- Position $s=t^3-6t^2+9t$: $v=3t^2-12t+9=3(t-1)(t-3)$, $a=6t-12$.
- At $v=0$ ($t=1,3$): $a=-6$ and $a=6$. At $a=0$ ($t=2$): $v=-3$, speed $3$.
- Distance from $0$ to $2$ is $4+2=6$ (forward then back), the displacement is $2$.
- Population $b=10^6+10^4t-10^3t^2$: growth rate $10^4-2000t$: $10^4,\ 0,\ -10^4$ at $t=0,5,10$.
Typical mistakes
Distance vs displacementSplit the motion where $v$ changes sign.
Continuity ≠ differentiabilityA corner is continuous but has no derivative.
Using $f'(x)$ at a junctionAt a junction use the limit definition from each side.
Practice next
- Chapter quizzes Ch. 2 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Plot and derivative
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Differentiability (Computation problems 3, Q9(b))
For $f(x)=x^2\sin\frac1x$ ($x\ne0$), $f(0)=0$, what is $f'(0)$?
Show the answer and the solution
Answer: $0$
$\frac{f(h)-0}{h}=h\sin\frac1h$ and $|h\sin\frac1h|\le|h|\to0$: $f'(0)=0$.
Worked example 2 · Rates of change (Tutorial 4, Q2(c))
For $s=t^3-6t^2+9t$, the total distance travelled from $t=0$ to $t=2$ is
Show the answer and the solution
Answer: $6$
The body moves forward until $t=1$ ($s$ goes $0\to4$) then back ($4\to2$): $4+2=6$ (the displacement is only $2$).
Worked example 3 · Differentiability (Computation problems 4, Q13)
For $h(x)=x^2\sin\frac1x$, $h(0)=0$: is $h'$ continuous at $0$? And for $k(x)=xh(x)$?
Show the answer and the solution
Answer: No for $h'$; yes for $k'$
$h'(x)=2x\sin\frac1x-\cos\frac1x$ for $x\ne0$ has no limit at $0$ although $h'(0)=0$. For $k=x^3\sin\frac1x$: $k'(x)=3x^2\sin\frac1x-x\cos\frac1x\to0=k'(0)$.
Worked example 4 · Rates of change (Tutorial 4, Q4)
An aircraft's distance along the runway is $D=\frac{10}{9}t^2$ m ($t$ in s). It takes off at $200$ km/h $=\frac{500}{9}$ m/s. After how many seconds?
Show the answer and the solution
Answer: $25$
$v=D'=\frac{20}{9}t=\frac{500}{9}\Rightarrow t=25$ s (distance $\frac{10}{9}\cdot625\approx694.4$ m).
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Plot and derivative