Exercise 1: The derivative at a point by the definition: brackets first

Thomas defines the slope of the curve $y = f(x)$ at $P(a, f(a))$ , and the derivative of $f$ at $a$ , by the same limit: $f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$ , or equivalently $f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ . The tangent at $P$ is the line through $P$ with slope $f'(a)$ .

Take $f(x) = 4 - 3x - x^2$ and $a = -1$ . The figure shows the parabola, the point $P(-1, 6)$ and three lines $L_1$ , $L_2$ , $L_3$ through $P$ . The differentiation rules of section 3.3 are NOT allowed as a method here.

a) Compute $f(-1)$ , then write $f(-1 + h)$ with brackets and expand it. Find $f'(-1)$ with the $h$ form of the definition.
b) Find $f'(-1)$ again with the $x \to a$ form, by factoring the numerator.
c) Write the equation of the tangent to the parabola at $P$ .
d) Which of the three lines of the figure is the tangent? Each of the two others meets the parabola at a second point: find its $x$ -coordinate.
e) Use the $x \to a$ form at a general point $a$ to find $f'(a)$ , then the point of the parabola where the tangent is horizontal.

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Answers

a) $f(-1) = 6$ , $f(-1 + h) = 6 - h - h^2$ , $f'(-1) = \lim_{h\to 0}(-1 - h) = -1$
b) $\frac{-x^2 - 3x - 2}{x + 1} = -(x + 2)$ for $x \ne -1$ , so $f'(-1) = -1$
c) $y = 5 - x$
d) $L_2$ is the tangent; $L_1$ ( $y = 3 - 3x$ ) meets the parabola again at $x = 1$ , $L_3$ ( $y = 7 + x$ ) at $x = -3$
e) $f'(a) = -3 - 2a$ ; horizontal tangent at $(-1.5, 6.25)$

a) $f(-1) = 4 + 3 - 1 = 6$ . Replace EVERY $x$ by $(-1 + h)$ , with brackets: $f(-1 + h) = 4 - 3(-1 + h) - (-1 + h)^2$ . The square is $(-1 + h)^2 = 1 - 2h + h^2$ , and the minus sign in front of it applies to ALL THREE terms: $-(1 - 2h + h^2) = -1 + 2h - h^2$ . So $f(-1 + h) = 4 + 3 - 3h - 1 + 2h - h^2 = 6 - h - h^2$ . The constants cancel in $f(-1 + h) - f(-1) = -h - h^2$ , as they always must. For $h \ne 0$ , $\frac{-h - h^2}{h} = -1 - h$ , and $f'(-1) = \lim_{h\to 0}(-1 - h) = -1$ . This is the gesture where MATH 203 marks go: the calculus is one line, the minus sign in front of a bracket is the whole difficulty.

b) $\frac{f(x) - f(-1)}{x - (-1)} = \frac{4 - 3x - x^2 - 6}{x + 1} = \frac{-x^2 - 3x - 2}{x + 1}$ . Factor out the minus sign first: $-x^2 - 3x - 2 = -(x^2 + 3x + 2) = -(x + 1)(x + 2)$ . For $x \ne -1$ the quotient is $-(x + 2)$ , and $f'(-1) = \lim_{x\to -1} -(x + 2) = -1$ , as in a). The root test is the safety net of this form: $x = -1$ always makes the numerator $0$ , so $(x + 1)$ always divides it. If it does not, $f(-1)$ was computed wrongly.

c) The tangent passes through $P(-1, 6)$ with slope $-1$ : $y - 6 = -1\,(x + 1)$ , that is $y = 5 - x$ . Check: $x = -1$ gives $y = 6$ .

d) $L_2$ has slope $-1$ (from $(-3, 8)$ to $(2, 3)$ on the grid): it is the tangent $y = 5 - x$ , and $4 - 3x - x^2 = 5 - x$ gives $x^2 + 2x + 1 = (x + 1)^2 = 0$ , a DOUBLE root, the algebraic signature of a tangent. $L_1$ is $y = 3 - 3x$ , slope $-3$ : $4 - 3x - x^2 = 3 - 3x$ gives $x^2 = 1$ , so it meets the curve again at $x = 1$ , the point $(1, 0)$ . $L_1$ is the secant $PQ$ with $Q(1, 0)$ , and slope $-3$ is exactly what a student gets by writing $(-1 + h)^2 = 1 + h^2$ : the middle term $-2h$ lost, the answer is the slope of a secant. $L_3$ is $y = 7 + x$ , slope $1$ : $4 - 3x - x^2 = 7 + x$ gives $x^2 + 4x + 3 = (x + 1)(x + 3) = 0$ , second point $x = -3$ , that is $(-3, 4)$ . Every line through $P$ other than the tangent cuts the parabola a second time; only the tangent gives a double root. A student who distributes the minus sign to the first term only, $-(-1 + h)^2 = -1 - 2h + h^2$ , gets $f(-1 + h) = 6 - 5h + h^2$ and the slope $-5$ of yet another secant.

e) $f(x) - f(a) = -3(x - a) - (x^2 - a^2) = -3(x - a) - (x - a)(x + a) = -(x - a)(3 + x + a)$ . For $x \ne a$ the quotient is $-(3 + x + a)$ , so $f'(a) = -(3 + 2a) = -3 - 2a$ . Check: $a = -1$ gives $-1$ . A horizontal tangent needs $f'(a) = 0$ , so $a = -1.5$ , and $f(-1.5) = 4 + 4.5 - 2.25 = 6.25$ : the vertex $(-1.5, 6.25)$ of the parabola. The difference of squares $x^2 - a^2 = (x - a)(x + a)$ is the one factoring the $x \to a$ form uses all the time.

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Exercise 2: Roots and fractions: the conjugate and the complex fraction

With a root or a fraction, the difference quotient does not simplify by expanding. Two gestures do the work. A ROOT: multiply top and bottom by the conjugate, and leave the bottom factored. A FRACTION: the numerator $f(a + h) - f(a)$ is a difference of fractions, so it is put over ONE common denominator first; the resulting numerator must be a multiple of $h$ , otherwise an error has been made. Dividing by $h$ then means multiplying the denominator by $h$ .

Use the definition only. A calculator is allowed, but it cannot compute a limit: it can only check the final number.

a) $g(x) = \sqrt{3x + 4}$ : find $g'(4)$ .
b) $k(x) = \frac{x + 1}{x - 1}$ : find $k'(3)$ with the $h$ form.
c) $m(x) = \frac{4}{\sqrt{x + 12}}$ : find $m'(4)$ . Both gestures are needed, in this order: common denominator, then conjugate.
d) Write the equation of the tangent to $y = k(x)$ at $x = 3$ .
e) A student computing b) writes $\frac{\frac{4+h}{2+h} - 2}{h} = \frac{4 + h - 2}{h(2 + h)} = \frac{1}{h}$ and concludes that $k'(3)$ does not exist. Find the error.

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Answers

a) $\frac{\sqrt{16 + 3h} - 4}{h} = \frac{3}{\sqrt{16 + 3h} + 4}$ , so $g'(4) = \frac{3}{8}$
b) $\frac{k(3+h) - k(3)}{h} = \frac{-1}{2 + h}$ , so $k'(3) = -\frac{1}{2}$
c) $\frac{m(4+h) - m(4)}{h} = \frac{-1}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})}$ , so $m'(4) = -\frac{1}{32}$
d) $y = 2 - \frac{1}{2}(x - 3)$ , that is $y = -\frac{1}{2}x + \frac{7}{2}$
e) The $2$ was not multiplied by the common denominator: $2 = \frac{2(2+h)}{2+h}$ , and the numerator is $4 + h - 4 - 2h = -h$ .

a) $g(4) = \sqrt{16} = 4$ and $g(4 + h) = \sqrt{3(4 + h) + 4} = \sqrt{16 + 3h}$ . The quotient $\frac{\sqrt{16 + 3h} - 4}{h}$ is a $\frac{0}{0}$ form with a root: multiply by the conjugate $\sqrt{16 + 3h} + 4$ . The numerator becomes $(16 + 3h) - 16 = 3h$ , so for $h \ne 0$ the quotient is $\frac{3h}{h(\sqrt{16 + 3h} + 4)} = \frac{3}{\sqrt{16 + 3h} + 4}$ , and $g'(4) = \frac{3}{4 + 4} = \frac{3}{8}$ . Two slips cost marks here: writing $g(4 + h) = \sqrt{3 \cdot 4 + 4} + h$ , with $h$ outside the root, and multiplying out the conjugate in the denominator, which destroys the factor that is about to become $8$ .

b) $k(3) = \frac{4}{2} = 2$ and $k(3 + h) = \frac{4 + h}{2 + h}$ . Common denominator: $\frac{4 + h}{2 + h} - 2 = \frac{4 + h - 2(2 + h)}{2 + h} = \frac{4 + h - 4 - 2h}{2 + h} = \frac{-h}{2 + h}$ . The numerator is a multiple of $h$ , which is the sign that the algebra is right. Dividing by $h$ multiplies the denominator by $h$ : $\frac{-h}{h(2 + h)} = \frac{-1}{2 + h}$ for $h \ne 0$ , so $k'(3) = -\frac{1}{2}$ . The negative sign agrees with the graph: $k(x) = 1 + \frac{2}{x - 1}$ decreases for $x > 1$ .

c) $m(4) = \frac{4}{\sqrt{16}} = 1$ and $m(4 + h) = \frac{4}{\sqrt{16 + h}}$ . First gesture, common denominator: $\frac{4}{\sqrt{16 + h}} - 1 = \frac{4 - \sqrt{16 + h}}{\sqrt{16 + h}}$ . The numerator still has a root and still tends to $0$ : second gesture, the conjugate $4 + \sqrt{16 + h}$ , which gives $\frac{16 - (16 + h)}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})} = \frac{-h}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})}$ . Divide by $h$ : the quotient is $\frac{-1}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})}$ , which tends to $\frac{-1}{4 \times 8} = -\frac{1}{32}$ . Doing the conjugate first, on the whole fraction, is legal but produces a fraction of fractions three storeys high; the order common denominator, then conjugate, keeps every line readable. Check with the calculator: $\frac{m(4.001) - m(4)}{0.001} \approx -0.03125$ .

d) Through $(3, k(3)) = (3, 2)$ with slope $-\frac{1}{2}$ : $y - 2 = -\frac{1}{2}(x - 3)$ , so $y = -\frac{1}{2}x + \frac{7}{2}$ .

e) In $\frac{4 + h}{2 + h} - 2$ , the $2$ must be written over the denominator $2 + h$ before subtracting: $2 = \frac{2(2 + h)}{2 + h} = \frac{4 + 2h}{2 + h}$ . The student subtracted the bare $2$ from the numerator, as if $2 = \frac{2}{2 + h}$ . Correct numerator: $4 + h - 4 - 2h = -h$ , not $2 + h$ . The student's line had a warning in it: the numerator $2 + h$ does not tend to $0$ , and a difference quotient ALWAYS has a numerator tending to $0$ when $f$ is continuous at $a$ . A quotient that blows up like $\frac{1}{h}$ for a function as smooth as $k$ near $3$ is an algebra error, not a discovery.

Exercise 3: The derivative as a function, by the definition

When the point varies, the derivative becomes a function: $f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$ , written also $\frac{dy}{dx}$ or $\frac{d}{dx}f(x)$ . Its domain is the set of $x$ where the limit exists, possibly smaller than the domain of $f$ . The algebra is that of Exercise 2 with the letter $x$ kept: the common denominator now contains $x$ , and the numerator must still collapse to a multiple of $h$ .

Three functions: $p(x) = \frac{1}{x^2}$ , $q(x) = \sqrt{x^2 + 9}$ and $r(x) = \frac{2x - 1}{x + 3}$ .

a) Find $p'(x)$ by the definition, then $p'(2)$ .
b) Find $q'(x)$ by the definition, then $q'(4)$ .
c) Find $r'(x)$ by the definition, give the domain of $r'$ , and compute $r'(-1)$ .
d) Find the point of the curve $y = \frac{1}{x^2}$ where the tangent has slope $16$ .
e) Find the points of the curve $y = r(x)$ where the tangent is parallel to the line $y = 7x$ .

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Answers

a) $p'(x) = -\frac{2}{x^3}$ , $p'(2) = -\frac{1}{4}$
b) $q'(x) = \frac{x}{\sqrt{x^2 + 9}}$ , $q'(4) = \frac{4}{5}$
c) $r'(x) = \frac{7}{(x + 3)^2}$ for $x \ne -3$ ; $r'(-1) = \frac{7}{4}$
d) $x = -\frac{1}{2}$ , the point $\left(-\frac{1}{2}, 4\right)$
e) $(x + 3)^2 = 1$ : the points $(-2, -5)$ and $(-4, 9)$

a) Common denominator: $\frac{1}{(x + h)^2} - \frac{1}{x^2} = \frac{x^2 - (x + h)^2}{x^2 (x + h)^2}$ . The numerator, with the minus sign distributed over the whole square, is $x^2 - x^2 - 2xh - h^2 = -2xh - h^2 = -h(2x + h)$ . Dividing by $h \ne 0$ : $\frac{-(2x + h)}{x^2 (x + h)^2}$ , which tends to $\frac{-2x}{x^4} = -\frac{2}{x^3}$ for $x \ne 0$ . So $p'(x) = -\frac{2}{x^3}$ and $p'(2) = -\frac{2}{8} = -\frac{1}{4}$ . The typical wrong numerator is $x^2 - x^2 - 2xh + h^2$ : the minus sign reached the first two terms only, and the answer is still $-\frac{2}{x^3}$ at the limit by luck, which is why the SIGN must be checked on the line, not on the result.

b) $\frac{\sqrt{(x + h)^2 + 9} - \sqrt{x^2 + 9}}{h}$ : conjugate. The numerator becomes $(x + h)^2 + 9 - (x^2 + 9) = 2xh + h^2 = h(2x + h)$ , so for $h \ne 0$ the quotient is $\frac{2x + h}{\sqrt{(x + h)^2 + 9} + \sqrt{x^2 + 9}}$ , which tends to $\frac{2x}{2\sqrt{x^2 + 9}} = \frac{x}{\sqrt{x^2 + 9}}$ . Since $x^2 + 9 \ge 9 > 0$ , this exists for every real $x$ : the domain of $q'$ is all of $\mathbb{R}$ . And $q'(4) = \frac{4}{\sqrt{25}} = \frac{4}{5}$ . The slip to avoid: $\sqrt{(x + h)^2 + 9} = \sqrt{x^2 + 9} + h$ , which is false (try $x = 0$ , $h = 4$ : $5 \ne 7$ ).

c) $r(x + h) - r(x) = \frac{(2x + 2h - 1)(x + 3) - (2x - 1)(x + h + 3)}{(x + h + 3)(x + 3)}$ . Expand each product in full: $(2x + 2h - 1)(x + 3) = 2x^2 + 5x - 3 + 2hx + 6h$ and $(2x - 1)(x + h + 3) = 2x^2 + 5x - 3 + 2hx - h$ . Their difference is $6h - (-h) = 7h$ : everything else cancels, as it must. So $\frac{r(x + h) - r(x)}{h} = \frac{7}{(x + h + 3)(x + 3)} \to \frac{7}{(x + 3)^2}$ . The domain of $r'$ is that of $r$ , all $x \ne -3$ , and $r'(-1) = \frac{7}{4}$ . Subtracting the second product with the minus sign on its first three terms only, $- 2x^2 - 5x + 3 + 2hx - h$ , leaves $4hx + 5h$ in place of $7h$ in the numerator: the missing brackets change the answer.

d) Slope $16$ : $-\frac{2}{x^3} = 16$ , so $x^3 = -\frac{1}{8}$ and $x = -\frac{1}{2}$ . Then $p\left(-\frac{1}{2}\right) = \frac{1}{1/4} = 4$ : the point $\left(-\frac{1}{2}, 4\right)$ . Only one point: a cube has one real cube root. A positive slope for $x < 0$ agrees with the graph, which rises toward the asymptote $x = 0$ on the left.

e) The line $y = 7x$ has slope $7$ , so $\frac{7}{(x + 3)^2} = 7$ , $(x + 3)^2 = 1$ , $x + 3 = \pm 1$ , $x = -2$ or $x = -4$ . Then $r(-2) = \frac{-5}{1} = -5$ and $r(-4) = \frac{-9}{-1} = 9$ : the points $(-2, -5)$ and $(-4, 9)$ . Taking only the positive square root, $x + 3 = 1$ , loses the second point, and it is a full part of the answer.

Exercise 4: Recognizing a limit as a derivative

Read from right to left, the definition says that every limit of the shape $\lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$ or $\lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ IS a derivative. To recognize it, name $f$ and $a$ , then CHECK that the number subtracted is $f(a)$ and that the denominator is exactly the increment.

Four limits: (i) $\lim_{h\to 0} \frac{\frac{1}{4 + h} - \frac{1}{4}}{h}$ , (ii) $\lim_{x\to 3} \frac{x^4 - 81}{x - 3}$ , (iii) $\lim_{\theta\to \pi/3} \frac{\cos\theta - \frac{1}{2}}{\theta - \frac{\pi}{3}}$ , (iv) $\lim_{t\to -2} \frac{t^3 + 3t + 14}{t + 2}$ .

a) For each limit, give a function $f$ and a number $a$ such that the limit equals $f'(a)$ .
b) Compute limit (i) by algebra alone.
c) Compute limits (ii) and (iv) by algebra alone.
d) $f$ is differentiable at $2$ with $f'(2) = -3$ . Find $\lim_{h\to 0} \frac{f(2 - 3h) - f(2)}{h}$ and $\lim_{x\to 2} \frac{f(x) - f(2)}{x^2 - 4}$ .
e) With the same $f$ , find $\lim_{h\to 0} \frac{f(2 + h) - f(2 - h)}{h}$ . Then compute this limit for $F(x) = |x - 2|$ , and explain why it does not prove that $F'(2)$ exists.

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Answers

a) (i) $\frac{1}{x}$ at $4$ ; (ii) $x^4$ at $3$ ; (iii) $\cos\theta$ at $\frac{\pi}{3}$ ; (iv) $t^3 + 3t$ at $-2$
b) $-\frac{1}{16}$
c) (ii) $108$ ; (iv) $15$
d) $9$ and $-\frac{3}{4}$
e) $2f'(2) = -6$ ; for $|x - 2|$ the limit is $0$ , yet $F'(2)$ does not exist (corner).

a) (i) $\frac{1}{4} = f(4)$ for $f(x) = \frac{1}{x}$ , and the denominator is the increment $h$ : $f'(4)$ . (ii) $81 = 3^4$ : $f(x) = x^4$ , $a = 3$ , in the $x \to a$ form. (iii) $\frac{1}{2} = \cos\frac{\pi}{3}$ : $f(\theta) = \cos\theta$ , $a = \frac{\pi}{3}$ ; its VALUE needs the derivative of cosine, which is section 3.5, and the question asks only for $f$ and $a$ . Answering $a = \frac{1}{2}$ confuses the point with the value. (iv) The constant has to be read inside the numerator: with $f(t) = t^3 + 3t$ , $f(-2) = -8 - 6 = -14$ , so $f(t) - f(-2) = t^3 + 3t + 14$ and $a = -2$ . Writing the line $f(-2) = -14$ is the proof that the shape really is $f(t) - f(a)$ .

b) Complex fraction: $\frac{1}{4 + h} - \frac{1}{4} = \frac{4 - (4 + h)}{4(4 + h)} = \frac{-h}{4(4 + h)}$ . Dividing by $h \ne 0$ multiplies the denominator by $h$ : $\frac{-h}{4h(4 + h)} = \frac{-1}{4(4 + h)}$ , which tends to $-\frac{1}{16}$ . The bracket in $4 - (4 + h)$ carries the sign of the answer: without it the numerator is $4 - 4 + h = h$ and the result comes out positive, although $\frac{1}{x}$ decreases.

c) (ii) Difference of squares twice: $x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$ . For $x \ne 3$ the quotient is $(x + 3)(x^2 + 9)$ , which tends to $6 \times 18 = 108$ . (iv) $t = -2$ is a root of $t^3 + 3t + 14$ , so $t + 2$ divides it: $t^3 + 3t + 14 = (t + 2)(t^2 - 2t + 7)$ , as expanding confirms ( $t^3 - 2t^2 + 7t + 2t^2 - 4t + 14$ ). The limit is $4 + 4 + 7 = 15$ . The rules of section 3.3 will check both in one line: $4 \cdot 27 = 108$ and $3 \cdot 4 + 3 = 15$ .

d) Put $k = -3h$ : as $h \to 0$ , $k \to 0$ , and $\frac{f(2 - 3h) - f(2)}{h} = -3 \cdot \frac{f(2 + k) - f(2)}{k} \to -3f'(2) = 9$ . For the second: $\frac{f(x) - f(2)}{x^2 - 4} = \frac{f(x) - f(2)}{x - 2} \cdot \frac{1}{x + 2}$ for $x \ne 2$ , and the product law gives $f'(2) \cdot \frac{1}{4} = -\frac{3}{4}$ . In both cases the quotient was REWRITTEN until the true difference quotient appeared, times a factor whose limit is known. Answering $-3$ twice, because both look like derivatives, is the trap.

e) Split the numerator by adding and subtracting $f(2)$ : $\frac{f(2 + h) - f(2 - h)}{h} = \frac{f(2 + h) - f(2)}{h} + \frac{f(2 - h) - f(2)}{-h}$ . Each term tends to $f'(2)$ (the second with increment $-h$ ), so the limit is $2f'(2) = -6$ . For $F(x) = |x - 2|$ : $F(2 + h) - F(2 - h) = |h| - |-h| = 0$ , so the symmetric quotient is $0$ for every $h$ , and its limit is $0$ . Yet $F$ has a corner at $2$ , one-sided slopes $1$ and $-1$ , and $F'(2)$ does not exist. The symmetric quotient averages the two one-sided slopes; it can exist when they differ, so it is NOT a definition of the derivative. Only the quotient with $f(a)$ in it is.

Exercise 5: Corner, cusp, jump, or a smooth join: where is g differentiable?

$f$ is differentiable at $a$ when the two one-sided limits of $\frac{f(a + h) - f(a)}{h}$ , as $h \to 0^-$ and $h \to 0^+$ , exist, are FINITE and are EQUAL. Thomas lists where this fails: a corner, a cusp, a vertical tangent, a discontinuity. A change of formula is NOT on the list: two pieces can join smoothly.

The figure shows $g$ on $[-4, 6.5]$ , given by $g(x) = x + 4$ on $[-4, -3]$ , $g(x) = 3 - 2\sqrt{|x + 2|}$ on $[-3, -1]$ , $g(x) = \frac{(x - 1)^2}{4}$ on $[-1, 3]$ , $g(x) = 1 - \frac{x - 3}{2}$ on $[3, 5)$ and $g(x) = 2$ on $[5, 6.5]$ . The open dot is not on the graph.

a) From the figure, where in $(-4, 6.5)$ is $g$ not continuous? Where is $g$ continuous but not differentiable, and what is the defect at each such point?
b) Compute the two one-sided derivatives of $g$ at $x = 3$ from the definition, and conclude.
c) The formula of $g$ also changes at $x = -1$ . Compute both one-sided derivatives there from the definition, and conclude.
d) $s(x) = |\sin x|$ : using the theorem $\lim_{\theta\to 0} \frac{\sin\theta}{\theta} = 1$ , compute the one-sided derivatives of $s$ at $0$ and conclude.
e) $u(x) = x^{4/3}$ and $v(x) = x^{1/5}$ both have fractional exponents. Study the difference quotient of each at $0$ and conclude.

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Answers

a) Not continuous at $5$ (jump). Continuous, not differentiable: $-2$ (cusp), $3$ (corner).
b) Left: $1$ ; right: $-\frac{1}{2}$ . Corner: $g'(3)$ does not exist.
c) Left: $-1$ ; right: $-1$ . $g$ is differentiable at $-1$ with $g'(-1) = -1$ .
d) Right: $1$ ; left: $-1$ . Corner: $s'(0)$ does not exist.
e) $\frac{u(h)}{h} = h^{1/3} \to 0$ : $u'(0) = 0$ . $\frac{v(h)}{h} = h^{-4/5} \to +\infty$ : vertical tangent, no $v'(0)$ .

a) At $x = 5$ the graph arrives at the open dot $(5, 0)$ while $g(5) = 2$ : $\lim_{x\to 5^-} g(x) = 0 \ne g(5)$ , a jump, so no derivative either. At $x = -2$ the graph comes UP to the point $(-2, 3)$ steeply from both sides and goes down again: the secants from the left become vertical upward, those from the right vertical downward, a CUSP. At $x = 3$ the curve arrives with one slope and leaves with another: a CORNER. At $x = 1$ the curve has its lowest point with a horizontal tangent: that is a perfectly good derivative, $g'(1) = 0$ , not a defect. At $x = -3$ and $x = -1$ the formula changes, but the figure shows no break in direction; c) checks $-1$ by computation.

b) $g(3) = \frac{(3 - 1)^2}{4} = 1$ , from the formula valid AT $3$ . For $h < 0$ , $3 + h$ is on the parabola: $\frac{\frac{(2 + h)^2}{4} - 1}{h} = \frac{(2 + h)^2 - 4}{4h} = \frac{4h + h^2}{4h} = 1 + \frac{h}{4} \to 1$ . For $h > 0$ , $3 + h$ is on the line: $\frac{1 - \frac{h}{2} - 1}{h} = -\frac{1}{2}$ . The one-sided derivatives, $1$ and $-\frac{1}{2}$ , are finite and different: a corner, $g'(3)$ does not exist. The complex fraction $\frac{\frac{(2+h)^2}{4} - 1}{h}$ is cleared by writing $1 = \frac{4}{4}$ and then multiplying the denominator by $h$ .

c) $g(-1) = \frac{(-2)^2}{4} = 1$ . For $h > 0$ , on the parabola: $\frac{\frac{(h - 2)^2}{4} - 1}{h} = \frac{h^2 - 4h}{4h} = \frac{h}{4} - 1 \to -1$ . For $h < 0$ (small), $-1 + h$ is on the middle piece, where $x + 2 = 1 + h > 0$ : $\frac{3 - 2\sqrt{1 + h} - 1}{h} = \frac{2(1 - \sqrt{1 + h})}{h}$ . Conjugate: $\frac{2(1 - (1 + h))}{h(1 + \sqrt{1 + h})} = \frac{-2}{1 + \sqrt{1 + h}} \to -1$ . Both one-sided derivatives equal $-1$ : $g$ IS differentiable at $-1$ , $g'(-1) = -1$ , although its formula changes there. A joint is a question, never an answer: the quotients decide. (The same computation at $-3$ gives $1$ on both sides.)

d) $s(0) = 0$ . For $0 < h < \pi$ , $\sin h > 0$ and $\frac{s(h) - s(0)}{h} = \frac{\sin h}{h} \to 1$ . For $-\pi < h < 0$ , $\sin h < 0$ , so $|\sin h| = -\sin h$ and the quotient is $-\frac{\sin h}{h} \to -1$ . The one-sided derivatives $1$ and $-1$ differ: a corner at the origin, exactly like $|x|$ . The absolute value is removed by cases BEFORE the limit, with the sign of $\sin h$ checked on each side. The theorem on $\frac{\sin\theta}{\theta}$ is used as a TOOL; no derivative of sine is needed.

e) $u(0) = 0$ and $\frac{u(h)}{h} = \frac{(h^{1/3})^4}{h} = h^{1/3}$ for $h \ne 0$ , which tends to $0$ from both sides: $u$ is differentiable at $0$ and $u'(0) = 0$ , with a horizontal tangent. $v(0) = 0$ and $\frac{v(h)}{h} = \frac{h^{1/5}}{h} = h^{-4/5} = \frac{1}{(h^{1/5})^4}$ , positive on both sides and tending to $+\infty$ : the tangent at the origin is the vertical line $x = 0$ , and $v'(0)$ does not exist. The exponent rule is the algebra that decides: $\frac{h^{p}}{h} = h^{p - 1}$ , and the quotient tends to $0$ when $p - 1 > 0$ , blows up when $p - 1 < 0$ . A fractional exponent is not by itself a defect.

Exercise 6: Graphing f' from the graph of f

Thomas 3.2 graphs the derivative by reading slopes: at each $x$ , $f'(x)$ is the slope of the tangent to the graph of $f$ . Where $f$ climbs, $f'$ is above the axis; where $f$ falls, below; a horizontal tangent puts $f'$ on the axis; a corner leaves $f'$ undefined. The HEIGHT of $f$ plays no role, only its slope.

The figure shows $f$ on $[-4, 5]$ : a smooth arc from $(-4, -2)$ to $(2, -2)$ , symmetric about the origin, with a horizontal tangent at its top $(-2, 2)$ and arriving flat at $(2, -2)$ ; then a segment from $(2, -2)$ to $(5, 1)$ . The dashed line is the tangent at the origin.

a) Read $f'(-2)$ , $f'(0)$ and $f'(4)$ on the figure.
b) Give the two one-sided derivatives of $f$ at $x = 2$ , and conclude.
c) Give the intervals where $f' > 0$ and where $f' < 0$ , and the number of points of $(-4, 5)$ where $f' = 0$ .
d) Sketch the graph of $f'$ on $(-4, 5)$ . Where on $(-2, 2)$ is $f'$ smallest, and what is its value on $(2, 5)$ ?
e) A classmate draws $f'$ as one unbroken curve, going through $(2, 0)$ and then jumping up to $1$ with a vertical stroke. What is $f'(2)$ really?

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Answers

a) $f'(-2) = 0$ , $f'(0) = -1.5$ , $f'(4) = 1$
b) Left: $0$ ; right: $1$ . Corner: $f'(2)$ does not exist.
c) $f' > 0$ on $(-4, -2)$ and $(2, 5)$ ; $f' < 0$ on $(-2, 2)$ ; $f' = 0$ at $x = -2$ only.
d) A U-shaped arc on $(-4, 2)$ through $(-2, 0)$ , lowest at $(0, -1.5)$ , reaching $0$ as $x \to 2^-$ ; then the constant $1$ on $(2, 5)$ , open dots at $(2, 0)$ and $(2, 1)$ .
e) $f'(2)$ does not exist: nothing is plotted at $x = 2$ , and a vertical stroke is never part of the graph of a function.

a) At $(-2, 2)$ the tangent is horizontal: $f'(-2) = 0$ . The dashed tangent at the origin passes through $(0, 0)$ and $(2, -3)$ on the grid, so $f'(0) = \frac{-3 - 0}{2 - 0} = -1.5$ . On the segment from $(2, -2)$ to $(5, 1)$ the slope is $\frac{1 - (-2)}{5 - 2} = 1$ , and the tangent to a line is the line itself: $f'(4) = 1$ . Reading a slope means reading TWO points of the tangent and dividing; reading the height $f(0) = 0$ answers another question.

b) From the left, the arc arrives at $(2, -2)$ with a horizontal tangent: the left derivative is $0$ . From the right, the segment leaves with slope $1$ : the right derivative is $1$ . Both are finite and they differ, so $f$ has a corner at $x = 2$ and $f'(2)$ does not exist, although $f$ is continuous there. The flat arrival from the left is the trap: it makes $(2, -2)$ look like a smooth bottom, and it is only half of one.

c) $f$ rises on $(-4, -2)$ and on $(2, 5)$ : $f' > 0$ there. $f$ falls on $(-2, 2)$ : $f' < 0$ there. $f' = 0$ only at $x = -2$ , ONE point: at $x = 2$ the derivative does not exist (b), so it cannot be $0$ . The sign of $f'$ is read as up or down, never as above or below the axis: $f$ is positive on $(-2, 0)$ while $f'$ is negative there.

d) On $(-4, 2)$ , $f'$ starts high (the arc is steep near $(-4, -2)$ ), decreases to $0$ at $x = -2$ , keeps decreasing to its lowest value at $x = 0$ , where the arc falls most steeply, $f'(0) = -1.5$ , then climbs back toward $0$ as $x \to 2^-$ . On $(2, 5)$ , $f' = 1$ , a horizontal segment. At $x = 2$ the graph of $f'$ has open dots at $(2, 0)$ and $(2, 1)$ and no filled dot. For the record, the arc is $y = \frac{x^3 - 12x}{8}$ , whose derivative, by the definition as in Exercise 3, is $\frac{3x^2 - 12}{8}$ : a parabola with vertex $(0, -1.5)$ and zeros $\pm 2$ , which is exactly the shape read from the slopes.

e) $f'(2)$ does not exist (b), so the graph of $f'$ has NO point above $x = 2$ : two open dots, $(2, 0)$ closing the left piece and $(2, 1)$ opening the right one. The classmate's vertical stroke would give $f'$ infinitely many values at $x = 2$ , which no function can have. The error comes from drawing $f'$ in one stroke because $f$ is drawn in one stroke: a continuous $f$ can have a derivative that jumps, and each corner of $f$ is a jump of $f'$ .

Exercise 7: Two parameters for a smooth join

Let $f(x) = ax^2 + b$ for $x < 2$ and $f(x) = \frac{12}{x}$ for $x \ge 2$ , where $a$ and $b$ are constants. The figure shows the right-hand piece only; the parabola on the left is to be chosen so that the graph has no break AND no corner at $(2, 6)$ .

Everything is computed from the definition, one side at a time. Remember that $f(2)$ comes from the formula valid AT $2$ , here $\frac{12}{x}$ , whichever side $2 + h$ is on.

a) Compute the right-hand derivative of $f$ at $2$ from the definition.
b) Write the condition on $a$ and $b$ for $f$ to be continuous at $2$ . Assuming it holds, show that the left-hand quotient equals $a(4 + h)$ for $h < 0$ . For $a = 1$ , which $b$ makes $f$ continuous, and what is then the left-hand derivative?
c) Find $a$ and $b$ so that $f$ is differentiable at $2$ .
d) A student proposes $a = -1$ and $b = 10$ . Is $f$ then continuous at $2$ ? Differentiable at $2$ ?
e) With the values of c), write the tangent at $(2, 6)$ and check that it meets each piece of the graph only at that point.

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Answers

a) $\frac{\frac{12}{2 + h} - 6}{h} = \frac{-6}{2 + h} \to -3$
b) $4a + b = 6$ ; for $a = 1$ : $b = 2$ , left derivative $4$
c) $4a = -3$ : $a = -\frac{3}{4}$ , $b = 9$
d) Continuous ( $-4 + 10 = 6$ ) but not differentiable: slopes $-4$ on the left, $-3$ on the right, a corner.
e) $y = 12 - 3x$ ; it meets $y = 9 - \frac{3}{4}x^2$ and $y = \frac{12}{x}$ only at $x = 2$ (double roots).

a) $f(2) = \frac{12}{2} = 6$ . For $h > 0$ : $\frac{12}{2 + h} - 6 = \frac{12 - 6(2 + h)}{2 + h} = \frac{12 - 12 - 6h}{2 + h} = \frac{-6h}{2 + h}$ . The numerator is a multiple of $h$ , as it must be. Divide by $h$ : $\frac{-6}{2 + h} \to -3$ . The right-hand derivative is $-3$ . The bracket in $6(2 + h)$ is where the marks go: $12 - 12 + h$ gives a positive slope for a decreasing curve.

b) $\lim_{x\to 2^-} f(x) = 4a + b$ and $f(2) = 6$ , so continuity at $2$ means $4a + b = 6$ . Then for $h < 0$ : $\frac{a(2 + h)^2 + b - 6}{h} = \frac{4a + 4ah + ah^2 + b - 6}{h} = \frac{4ah + ah^2}{h} = a(4 + h)$ , using $4a + b - 6 = 0$ . Its limit is $4a$ . Without continuity the constant $4a + b - 6$ would NOT cancel, the numerator would not tend to $0$ , and the quotient would blow up: this is why continuity comes first. For $a = 1$ : $b = 6 - 4 = 2$ , and the left derivative is $4$ , while the right one is $-3$ : continuous, with a corner.

c) Differentiable at $2$ : continuous, AND left derivative equal to right derivative, $4a = -3$ . So $a = -\frac{3}{4}$ and $b = 6 - 4a = 6 + 3 = 9$ . The left piece is $9 - \frac{3}{4}x^2$ , a downward parabola that meets the hyperbola at $(2, 6)$ with the same slope, as the figure of the solution shows. Two unknowns need two equations: continuity gives one, equal slopes the other. Matching only the slopes, $4a = -3$ , and forgetting $b$ leaves the graph broken.

d) $4a + b = -4 + 10 = 6$ : $f$ is continuous at $2$ . But the left derivative is $4a = -4$ and the right one is $-3$ : finite and different, so $f$ has a corner at $(2, 6)$ and is not differentiable there. Continuity is necessary for differentiability, never sufficient; the student checked the first condition only.

e) The tangent at $(2, 6)$ has slope $-3$ : $y = 6 - 3(x - 2) = 12 - 3x$ . With the parabola: $9 - \frac{3}{4}x^2 = 12 - 3x$ gives $\frac{3}{4}x^2 - 3x + 3 = \frac{3}{4}(x - 2)^2 = 0$ , only $x = 2$ . With the hyperbola: $\frac{12}{x} = 12 - 3x$ gives, for $x \ne 0$ , $12 = 12x - 3x^2$ , so $3x^2 - 12x + 12 = 3(x - 2)^2 = 0$ , only $x = 2$ . A double root on each side is the algebraic trace of a tangent, and the figure of the solution shows the line touching the join without crossing either piece.

Exercise 8: Five statements to correct

Each statement below comes from a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample with numbers, and write the correct statement.

a) If $f'(a) = 0$ , then the tangent to the graph of $f$ at $a$ is the $x$ -axis.
b) If $\lim_{h\to 0^+} \frac{f(a + h) - f(a)}{h}$ exists, then $f$ is differentiable at $a$ .
c) If $f(x) \le g(x)$ for every $x$ , then $f'(a) \le g'(a)$ for every $a$ .
d) $\frac{1}{x + h} - \frac{1}{x} = \frac{1}{h}$ , so the difference quotient of $\frac{1}{x}$ is $\frac{1}{h^2}$ and $\frac{1}{x}$ has no derivative.
e) $f(x) = x^2 - 5x$ gives $f(3) = -6$ , and the derivative of the constant $-6$ is $0$ , so $f'(3) = 0$ .

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Answers

a) False: $f(x) = (x - 1)^2 + 3$ has $f'(1) = 0$ and tangent $y = 3$ . The tangent is HORIZONTAL, $y = f(a)$ .
b) False: $|2x - 6|$ at $3$ has right-hand limit $2$ , left-hand limit $-2$ . Both one-sided limits must exist and be equal.
c) False: $0 \le x^2$ everywhere, yet at $a = -1$ the slopes are $0$ and $-2$ . An inequality between values says nothing about slopes.
d) False: at $x = 1$ , $h = 1$ , $\frac{1}{2} - 1 = -\frac{1}{2} \ne 1$ . In fact $\frac{1}{x+h} - \frac{1}{x} = \frac{-h}{x(x+h)}$ , and $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$ .
e) False: $f'(3)$ is the slope of $f$ at $3$ , not the derivative of the number $f(3)$ ; $f'(3) = 1$ .

a) FALSE. $f(x) = (x - 1)^2 + 3$ : at $a = 1$ , $\frac{f(1 + h) - f(1)}{h} = \frac{h^2}{h} = h \to 0$ , so $f'(1) = 0$ , and the tangent is $y = f(1) + 0 \cdot (x - 1)$ , that is $y = 3$ , three units above the $x$ -axis. Correct statement: if $f'(a) = 0$ , the tangent at $a$ is HORIZONTAL, the line $y = f(a)$ . The slope is $0$ , the height is $f(a)$ .

b) FALSE. $f(x) = |2x - 6|$ at $a = 3$ , where $f(3) = 0$ : for $h > 0$ , $\frac{|2h|}{h} = 2$ ; for $h < 0$ , $\frac{|2h|}{h} = -2$ . The right-hand limit exists (it is $2$ ), but the left-hand one is $-2$ : a corner, no derivative. Correct statement: $f$ is differentiable at an interior point $a$ if and only if BOTH one-sided limits exist, are finite and are equal. A single one-sided derivative is what Thomas uses only at an ENDPOINT of a closed interval.

c) FALSE. $f(x) = 0$ and $g(x) = x^2$ satisfy $f \le g$ everywhere. At $a = -1$ : $f'(-1) = 0$ , and $\frac{g(-1 + h) - g(-1)}{h} = \frac{-2h + h^2}{h} = -2 + h \to -2$ , so $g'(-1) = -2 < 0 = f'(-1)$ . A curve can lie above another and be falling while the other is flat. Correct statement: an inequality between the VALUES of two functions gives no inequality between their slopes.

d) FALSE, and it is the algebra slip of the chapter. Test with numbers: $x = 1$ , $h = 1$ gives $\frac{1}{2} - 1 = -\frac{1}{2}$ , not $\frac{1}{1} = 1$ . Fractions subtract over a common denominator: $\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}$ . The difference quotient is then $\frac{-1}{x(x + h)}$ , which tends to $-\frac{1}{x^2}$ for $x \ne 0$ ; at $x = 2$ it gives $-\frac{1}{4}$ . Correct statement: $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$ for $x \ne 0$ . A numerator that does not contain $h$ as a factor is always a sign of an error.

e) FALSE. $f'(3)$ is the derivative of the FUNCTION $f$ evaluated at $3$ , not the derivative of the NUMBER $f(3)$ . Evaluate after differentiating, never before: $f(3 + h) = 9 + 6h + h^2 - 15 - 5h = -6 + h + h^2$ , so $\frac{f(3 + h) - f(3)}{h} = 1 + h \to 1$ . Correct statement: $f'(3) = 1$ ; the notation $f'(3)$ means: compute the limit of the difference quotient of $f$ at $3$ . The derivative of a constant is $0$ , but $f$ is not a constant.

Exercise 9: An oven preheating: the rate of change in words, with its units

An oven is switched on at $t = 0$ . Its temperature, in degrees Celsius, is modelled by $T(t) = 200 - \frac{720}{t + 4}$ for $0 \le t \le 30$ , with $t$ in minutes. The figure shows $T$ , the tangent at $t = 4$ and the secant on $[4, 12]$ .

$T'(t)$ is measured in degrees Celsius PER MINUTE: it is a rate at an instant, never a temperature. A calculator is allowed; every value here is exact.

a) Compute $T(0)$ , $T(4)$ and $T(12)$ , and the average rate of change of $T$ on $[4, 12]$ , with its units.
b) Find $T'(t)$ from the definition, then $T'(4)$ and $T'(12)$ .
c) Write one sentence, with units, that says what $T'(4)$ means. Use it to predict $T(5)$ , and compare with the exact $T(5)$ .
d) At what time is the oven heating at $5$ degrees per minute? What is its temperature then?
e) Find $\lim_{t\to\infty} T'(t)$ and $\lim_{t\to\infty} T(t)$ for the formula. A user reads the first limit and says the oven ends up cold. Correct him.

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Answers

a) $T(0) = 20$ , $T(4) = 110$ , $T(12) = 155$ degrees; average rate $\frac{45}{8} = 5.625$ degrees per minute
b) $T'(t) = \frac{720}{(t + 4)^2}$ ; $T'(4) = 11.25$ and $T'(12) = 2.8125$ degrees per minute
c) At $t = 4$ min the temperature rises at $11.25$ degrees per minute; $T(5) \approx 121.25$ , exact $T(5) = 120$ .
d) $(t + 4)^2 = 144$ : $t = 8$ min, $T(8) = 140$ degrees
e) $T'(t) \to 0$ and $T(t) \to 200$ : the RATE vanishes, the oven settles near $200$ degrees.

a) $T(0) = 200 - \frac{720}{4} = 20$ degrees, room temperature. $T(4) = 200 - \frac{720}{8} = 110$ and $T(12) = 200 - \frac{720}{16} = 155$ degrees. Average rate on $[4, 12]$ : $\frac{155 - 110}{12 - 4} = \frac{45}{8} = 5.625$ degrees per minute, the slope of the dashed secant. The answer $45$ degrees is the CHANGE of temperature, not a rate.

b) The constant $200$ cancels in $T(t + h) - T(t)$ , which leaves $-\frac{720}{t + h + 4} + \frac{720}{t + 4} = 720\left(\frac{1}{t + 4} - \frac{1}{t + h + 4}\right)$ . Common denominator: $\frac{1}{t + 4} - \frac{1}{t + h + 4} = \frac{(t + h + 4) - (t + 4)}{(t + 4)(t + h + 4)} = \frac{h}{(t + 4)(t + h + 4)}$ . So $\frac{T(t + h) - T(t)}{h} = \frac{720}{(t + 4)(t + h + 4)}$ , which tends to $\frac{720}{(t + 4)^2}$ . Two minus signs met on the way, the one in front of the fraction and the one of the subtraction; taking the fractions in the order $\frac{1}{t+4} - \frac{1}{t+h+4}$ absorbs both at once. $T'(4) = \frac{720}{64} = 11.25$ and $T'(12) = \frac{720}{256} = 2.8125$ degrees per minute. Positive: the oven is heating, as the rising graph says.

c) At $t = 4$ minutes, the oven temperature is rising at $11.25$ degrees Celsius per minute. The sentence names the instant, the rate with its units, and the direction. If the rate stayed the same for one minute, $T(5) \approx 110 + 11.25 = 121.25$ degrees; the model gives $T(5) = 200 - \frac{720}{9} = 120$ degrees. The prediction overshoots by $1.25$ degree because the heating slows down: the graph bends below its tangent, visible on the figure.

d) $\frac{720}{(t + 4)^2} = 5$ gives $(t + 4)^2 = 144$ , so $t + 4 = 12$ (the root $t + 4 = -12$ is rejected, since $t \ge 0$ ), $t = 8$ minutes. Then $T(8) = 200 - \frac{720}{12} = 140$ degrees. The rate goes from $45$ degrees per minute at $t = 0$ to $5$ at $t = 8$ : the oven heats fast, then more and more slowly.

e) As $t \to \infty$ , $(t + 4)^2 \to \infty$ , so $T'(t) = \frac{720}{(t + 4)^2} \to 0$ , and $\frac{720}{t + 4} \to 0$ , so $T(t) \to 200$ . The user confused the RATE with the TEMPERATURE: $T'(t) \to 0$ says that the temperature stops CHANGING, not that it drops to zero. The oven settles near $200$ degrees, the thermostat setting, and there the heating rate is almost zero because the temperature is almost constant. (The model is stated for $t \le 30$ ; at $t = 30$ , $T'(30) = \frac{720}{1156} \approx 0.62$ degree per minute already.)

Exercise 10: A tank that drains: derivative, flow rate and the sign

A tank is emptied through a valve at the bottom. The volume of water left after $t$ minutes is $V(t) = 2(60 - t)^2$ litres, for $0 \le t \le 60$ . The figure shows $V$ and its tangent at $t = 10$ .

$V'(t)$ is in litres per minute. The FLOW RATE through the valve, the quantity a technician reads on a meter, is a positive number of litres per minute: it is $-V'(t)$ , since the volume decreases.

a) Compute $V(0)$ , $V(10)$ and $V(60)$ , and the average rate of change of $V$ over the whole draining, with units.
b) Find $V'(t)$ from the definition, then $V'(10)$ and $V'(50)$ .
c) Say in one sentence what $V'(10)$ means, and give the flow rate through the valve at $t = 10$ .
d) At what time is the flow rate equal to the average flow rate of the whole draining? How much water is left then?
e) The tank is only defined on $[0, 60]$ . Compute the one-sided derivatives $V'(0)$ (from the right) and $V'(60)$ (from the left), and say what the second one means physically.

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Answers

a) $V(0) = 7200$ , $V(10) = 5000$ , $V(60) = 0$ litres; average rate $-120$ litres per minute
b) $V'(t) = -4(60 - t) = 4t - 240$ ; $V'(10) = -200$ and $V'(50) = -40$ litres per minute
c) At $t = 10$ min the volume is decreasing at $200$ litres per minute: flow rate $200$ L/min.
d) $4(60 - t) = 120$ : $t = 30$ min, $V(30) = 1800$ litres left
e) $V'(0) = -240$ and $V'(60) = 0$ litres per minute: the flow dies down to zero as the tank empties.

a) $V(0) = 2 \times 3600 = 7200$ litres, $V(10) = 2 \times 2500 = 5000$ litres, $V(60) = 0$ : the tank is empty after one hour. Average rate over $[0, 60]$ : $\frac{0 - 7200}{60} = -120$ litres per minute. The minus sign says the volume decreases; it is part of the answer.

b) Write $u = 60 - t$ , so that $60 - (t + h) = u - h$ . Then $V(t + h) - V(t) = 2\left[(u - h)^2 - u^2\right] = 2\left[-2uh + h^2\right]$ . The minus sign in $60 - (t + h)$ is the one students lose: writing $60 - t + h$ reverses the sign of the answer. For $h \ne 0$ : $\frac{V(t + h) - V(t)}{h} = -4u + 2h \to -4u$ , so $V'(t) = -4(60 - t) = 4t - 240$ litres per minute. $V'(10) = -200$ and $V'(50) = -40$ .

c) At $t = 10$ minutes, the volume of water in the tank is decreasing at $200$ litres per minute. The flow rate through the valve is $-V'(10) = 200$ litres per minute. The tangent of the figure shows it: from $(10, 5000)$ it drops $6000$ litres in $30$ minutes, $200$ per minute. Saying 200 litres are left, or 200 litres have drained, reads the rate as an amount.

d) The average flow rate is $120$ litres per minute, so $4(60 - t) = 120$ , $60 - t = 30$ , $t = 30$ minutes, and $V(30) = 2 \times 900 = 1800$ litres are left. At that moment the flow is exactly the average one: faster before, slower after.

e) At the endpoints only one side exists. At $0$ , for $h > 0$ : $\frac{V(h) - V(0)}{h} = \frac{2(60 - h)^2 - 7200}{h} = \frac{-240h + 2h^2}{h} = -240 + 2h \to -240$ . At $60$ , for $h < 0$ : $\frac{V(60 + h) - V(60)}{h} = \frac{2h^2}{h} = 2h \to 0$ . So $V'(0) = -240$ and $V'(60) = 0$ litres per minute: the water leaves at $240$ litres per minute when the valve opens, and the flow slows to zero as the tank empties, which is why the graph arrives flat at $(60, 0)$ . Thomas calls $V$ differentiable on the closed interval $[0, 60]$ because these one-sided derivatives exist at the ends.

Corrected exercises: the derivative at a point and as a function (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: The derivative at a point by the definition: brackets first

Exercise 2: Roots and fractions: the conjugate and the complex fraction

Exercise 3: The derivative as a function, by the definition

Exercise 4: Recognizing a limit as a derivative

Exercise 5: Corner, cusp, jump, or a smooth join: where is g differentiable?

Part B: problems and reasoning (/50)

Exercise 6: Graphing f' from the graph of f

Exercise 7: Two parameters for a smooth join

Exercise 8: Five statements to correct

Exercise 9: An oven preheating: the rate of change in words, with its units

Exercise 10: A tank that drains: derivative, flow rate and the sign

See also

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