The Derivative: Definition and Interpretation

Step 1 — build $f(x+h)$.

$f(x+h)=4-3(x+h)^2=4-3\left(x^2+2xh+h^2\right)=4-3x^2-6xh-3h^2$

Step 2 — subtract and factor.

$f(x+h)-f(x)=\left(4-3x^2-6xh-3h^2\right)-\left(4-3x^2\right)=-6xh-3h^2=h(-6x-3h)$

Step 3 — divide (legal, since $h\ne0$ in the limit) and take the limit.

(a) The pattern is $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ with $f(x)=x^4$ and $a=3$ (indeed $f(3)=81$). Evaluate directly by expanding:

$(3+h)^4=81+108h+54h^2+12h^3+h^4$

(a) Units and meaning. $V$ is measured in litres and $t$ in minutes, so

$V'(t)\ \text{has units}\ \frac{\text{litres}}{\text{minute}}$

$V'(10)=-3.5$ litres per minute means that at the instant $t=10$ minutes the water is leaving the tank (the sign is negative, so $V$ is decreasing) at an instantaneous rate of $3.5$ litres per minute.

(b) Linear estimates. The tangent line at $t=10$ is

$L(t)=V(10)+V'(10)(t-10)=180-3.5(t-10)$

For $t=10.5$ (so $\Delta t=0.5$):

$V(10.5)\approx 180-3.5(0.5)=180-1.75=178.25\ \text{litres}$

For $t=12$ (so $\Delta t=2$):

$V(12)\approx 180-3.5(2)=180-7=173\ \text{litres}$

(c) The estimate for $V(10.5)$ is more trustworthy. The approximation $V(t)\approx V(10)+V'(10)(t-10)$ replaces the graph of $V$ by its tangent line at $t=10$, which is only close to the graph for small $\Delta t$. Over $0.5$ minutes the drainage rate has little time to change; over $2$ minutes it may change substantially (a draining tank typically empties more slowly as the water level drops, i.e. $V'$ increases towards $0$), so the tangent-line value drifts away from the true value.

Answers: (a) litres per minute; the volume is decreasing at $3.5$ L/min at $t=10$. (b) $V(10.5)\approx 178.25$ L, $V(12)\approx 173$ L. (c) The $t=10.5$ estimate, because the linear approximation is only reliable for small changes in $t$.

(a) Average rate of change on $[0,8]$.

$\frac{T(8)-T(0)}{8-0}=\frac{59-92}{8}=\frac{-33}{8}=-4.125\ ^\circ\text{C per minute}$

(b) Three estimates of $T'(4)$. Each is a difference quotient centred on, or ending at, $t=4$.

Backward difference (uses $t=2$ and $t=4$):

$\frac{T(4)-T(2)}{4-2}=\frac{71-80}{2}=\frac{-9}{2}=-4.5\ ^\circ\text{C/min}$

Forward difference (uses $t=4$ and $t=6$):

$\frac{T(6)-T(4)}{6-4}=\frac{64-71}{2}=\frac{-7}{2}=-3.5\ ^\circ\text{C/min}$

Symmetric difference (uses $t=2$ and $t=6$):

$\frac{T(6)-T(2)}{6-2}=\frac{64-80}{4}=\frac{-16}{4}=-4\ ^\circ\text{C/min}$

Note the symmetric estimate is exactly the average of the other two, $\frac12(-4.5-3.5)=-4$, and it is normally the most accurate because the two one-sided errors partly cancel. So $T'(4)\approx -4\ ^\circ$C per minute: at $t=4$ minutes the coffee is cooling at about $4$ degrees Celsius per minute.

(c) Sign of $T''$. Successive secant slopes over $[0,2],[2,4],[4,6],[6,8]$ are

$\frac{80-92}{2}=-6,\qquad \frac{71-80}{2}=-4.5,\qquad \frac{64-71}{2}=-3.5,\qquad \frac{59-64}{2}=-2.5$

These are increasing ($-6<-4.5<-3.5<-2.5$). Five data points cannot prove anything about $T''$ — infinitely many functions pass through the same five points — but a steadily increasing run of secant slopes is exactly what an increasing $T'$ looks like, so the data indicate that $T'$ is increasing and hence that $T''>0$ on $[0,8]$. (Strictly, $T'$ increasing forces only $T''\ge 0$ wherever $T''$ exists; the strict inequality is the natural reading here.) Physically: the coffee is still cooling ($T'<0$) but the cooling is slowing down as the coffee approaches room temperature; the graph of $T$ is concave up.

Answers: (a) $-4.125\ ^\circ$C/min. (b) $-4.5$, $-3.5$ and $-4\ ^\circ$C/min respectively; best estimate $T'(4)\approx-4\ ^\circ$C/min. (c) the data indicate $T''>0$ (graph concave up): the rate of cooling is becoming less negative, so cooling slows over time.

(a) Derivative from the definition.

(a) Derivative at $9$. Since $x-9=\left(\sqrt x-3\right)\left(\sqrt x+3\right)$ for $x\ge0$,

(a) Continuity at $2$. For any $x$, $\left|\,|x-2|\,\right|=|x-2|\to 0$ as $x\to 2$, so

$\lim_{x\to 2}f(x)=\lim_{x\to 2}|x-2|=0=|2-2|=f(2)$

All three requirements of continuity hold ($f(2)$ defined, the limit exists, they are equal), so $f$ is continuous at $2$.

(b) One-sided derivatives. With $a=2$,

$\frac{f(2+h)-f(2)}{h}=\frac{|(2+h)-2|-0}{h}=\frac{|h|}{h}$

If $h>0$ then $|h|=h$, so the quotient equals $1$:

$f'_+(2)=\lim_{h\to 0^+}\frac{|h|}{h}=\lim_{h\to 0^+}1=1$

If $h<0$ then $|h|=-h$, so the quotient equals $-1$:

$f'_-(2)=\lim_{h\to 0^-}\frac{|h|}{h}=\lim_{h\to 0^-}(-1)=-1$

Both one-sided derivatives exist and are finite, but $1\ne-1$, so the two-sided limit

$\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$

does not exist. Hence $f'(2)$ does not exist: the graph has a corner at $(2,0)$.

(c) Consequence. $f$ is continuous at $2$ but not differentiable at $2$, so the converse of the theorem "differentiable at $a$ $\Rightarrow$ continuous at $a$" is false. Continuity is a necessary but not a sufficient condition for differentiability.

(d) The derivative function. For $x>2$ we have $f(x)=x-2$, a line of slope $1$; for $x<2$ we have $f(x)=2-x$, a line of slope $-1$. (Each of these can be confirmed from the definition exactly as in part (b), since near such an $x$ the absolute value is given by a single linear formula.) Thus

$f'(x)=\begin{cases}-1,& x<2\\ 1,& x>2\end{cases}$

with $\text{dom}\,f'=(-\infty,2)\cup(2,\infty)$. Note $f'$ itself has a jump discontinuity at $x=2$, and $\text{dom}\,f'$ is strictly smaller than $\text{dom}\,f=(-\infty,\infty)$.

Answers: (a) continuous, both sides give $0$; (b) $f'_+(2)=1$, $f'_-(2)=-1$, so $f'(2)$ does not exist; (c) the converse of the theorem is false; (d) $f'(x)=-1$ for $x<2$, $f'(x)=1$ for $x>2$, domain $x\ne2$.

Step 1 — continuity is forced. If $f$ is differentiable at $1$ then $f$ is continuous at $1$. Now $f(1)=1^2=1$ and

$\lim_{x\to 1^-}f(x)=\lim_{x\to1^-}x^2=1,\qquad \lim_{x\to 1^+}f(x)=\lim_{x\to1^+}(ax+b)=a+b$

so continuity requires

$a+b=1\qquad\text{(i)}$

Step 2 — left-hand derivative. For $h<0$, $1+h<1$, so $f(1+h)=(1+h)^2$:

(a) $f(x)=x^{2/3}$, $f(0)=0$.

$\frac{f(0+h)-f(0)}{h}=\frac{h^{2/3}}{h}=h^{2/3-1}=h^{-1/3}=\frac{1}{h^{1/3}}$

As $h\to 0^+$: $h^{1/3}\to 0^+$, so $\dfrac{1}{h^{1/3}}\to+\infty$. As $h\to 0^-$: $h^{1/3}\to 0^-$ (the cube root of a negative number is negative), so $\dfrac{1}{h^{1/3}}\to-\infty$.

Both one-sided limits are infinite and of opposite sign. The derivative does not exist, and the graph has a cusp at the origin: the curve comes into $(0,0)$ with infinitely steep slope on both sides, but the two branches point in opposite directions (like the sharp tip of the graph of $y=x^{2/3}$).

(b) $g(x)=x^{1/3}$, $g(0)=0$.

$\frac{g(0+h)-g(0)}{h}=\frac{h^{1/3}}{h}=h^{1/3-1}=h^{-2/3}=\frac{1}{\left(h^{1/3}\right)^{2}}$

The denominator is a square, hence positive for every $h\ne 0$, and it tends to $0$. So

$\lim_{h\to 0^+}h^{-2/3}=+\infty\quad\text{and}\quad\lim_{h\to 0^-}h^{-2/3}=+\infty$

Both sides tend to $+\infty$, so $g'(0)$ does not exist, but the secant lines rotate to a single limiting position: the graph has a vertical tangent line $x=0$ at the origin. (Unlike (a), the curve passes smoothly through, merely infinitely steeply.)

(c) $k$. First check continuity: $k(0)=0+2=2$, while

$\lim_{x\to 0^-}k(x)=\lim_{x\to0^-}x=0\ne 2$

so $k$ has a jump discontinuity at $0$. By the contrapositive of "differentiable $\Rightarrow$ continuous", $k$ is not differentiable at $0$. Directly from the definition, for $h>0$,

$\frac{k(h)-k(0)}{h}=\frac{(h+2)-2}{h}=1\ \to\ 1$

but for $h<0$,

$\frac{k(h)-k(0)}{h}=\frac{h-2}{h}=1-\frac{2}{h}\ \to\ +\infty\quad (h\to 0^-,\ \text{since }-2/h>0\text{ and }\to+\infty)$

so the two-sided limit certainly fails to exist.

Answers: (a) cusp (quotient $\to+\infty$ on the right, $-\infty$ on the left); (b) vertical tangent $x=0$ (quotient $\to+\infty$ on both sides); (c) jump discontinuity, hence not differentiable. In none of the three cases is there a corner — a corner requires finite but unequal one-sided derivatives, as for $|x|$.

(a) First derivative.

Throughout we use only the bound $-1\le\sin\theta\le 1$, valid for every real $\theta$, together with the Squeeze Theorem. Note $\sin(1/x)$ has no limit as $x\to 0$, so no limit law may be applied to it directly.

(a) Continuity at $0$. For $x\ne 0$,

$-|x|\le x\sin\left(\frac1x\right)\le |x|$

Since $\lim_{x\to0}(-|x|)=0=\lim_{x\to0}|x|$, the Squeeze Theorem gives $\lim_{x\to0}g(x)=0=g(0)$, so $g$ is continuous at $0$. The identical argument with $-x^2\le x^2\sin(1/x)\le x^2$ gives $\lim_{x\to0}f(x)=0=f(0)$, so $f$ is continuous at $0$.

(b) $g$ is not differentiable at $0$. By the definition,

$\frac{g(0+h)-g(0)}{h}=\frac{h\sin\left(\frac1h\right)-0}{h}=\sin\left(\frac1h\right)\qquad(h\ne0)$

So the question is whether $\lim_{h\to0}\sin(1/h)$ exists — and it does not. Take two sequences tending to $0$:

$h_n=\frac{1}{2\pi n}\ \Rightarrow\ \sin\left(\frac{1}{h_n}\right)=\sin(2\pi n)=0$ $k_n=\frac{1}{\frac{\pi}{2}+2\pi n}\ \Rightarrow\ \sin\left(\frac{1}{k_n}\right)=\sin\left(\frac{\pi}{2}+2\pi n\right)=1$

Both $h_n\to 0$ and $k_n\to 0$, yet the difference quotient tends to $0$ along one sequence and to $1$ along the other. A limit, if it existed, would have to be unique, so no limit exists. Hence $g'(0)$ does not exist. This is not a corner, cusp or vertical tangent: the quotient stays bounded between $-1$ and $1$ but oscillates infinitely often.

(c) $f$ is differentiable at $0$.

$\frac{f(0+h)-f(0)}{h}=\frac{h^2\sin\left(\frac1h\right)-0}{h}=h\sin\left(\frac1h\right)\qquad(h\ne0)$

Now the extra factor of $h$ saves us:

$-|h|\le h\sin\left(\frac1h\right)\le|h|$

and $\lim_{h\to0}\left(-|h|\right)=\lim_{h\to0}|h|=0$, so by the Squeeze Theorem

$f'(0)=\lim_{h\to0}h\sin\left(\frac1h\right)=0$

The derivative exists and equals $0$: the tangent line at the origin is the horizontal line $y=0$, even though the graph crosses it infinitely often in every interval around $0$.

(d) Moral. Continuity at a point tells you nothing about differentiability there ($g$ is continuous but not differentiable), and failure of differentiability need not look like the standard pictures — there is no corner, cusp, vertical tangent or jump in $g$ at $0$, only infinitely fast oscillation. Conversely, differentiability is a genuinely stronger, more delicate condition than continuity: multiplying by one extra factor of $x$ is enough to turn the same oscillation from fatal into harmless.

Answers: (a) both continuous at $0$ by squeezing; (b) $g'(0)$ does not exist because $\sin(1/h)$ has no limit; (c) $f'(0)=0$ by the Squeeze Theorem; (d) continuity does not imply differentiability, and non-differentiability is not always a corner or cusp.

(a) Velocity from the definition. Expand: