Self-check: 201-SN2-RE, Differential Calculus

a) $(f\circ g)(x)=f\left(\sqrt{x}\right)=\frac{1}{\sqrt{x}-2}$. The domain is built in TWO steps, never in one. First step: $x$ must belong to the domain of $g$, so $x\ge 0$. Second step: $g(x)$ must belong to the domain of $f$, so $\sqrt{x}\neq 2$, that is $x\neq 4$. Conclusion: $\text{dom}(f\circ g)=\left[0, 4\right)\cup\left(4, +\infty\right)$. Reading the domain off the simplified expression $\frac{1}{\sqrt{x}-2}$ would give the right result here by luck, but the two-step method is the only safe one.

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a) General method: divide top and bottom by the highest power present, here $x^{2}$: $f(x)=\frac{3-\frac{2}{x}+\frac{1}{x^{2}}}{1-\frac{4}{x^{2}}}$. As $x\to\pm\infty$, all the terms in $\frac{1}{x}$ and $\frac{1}{x^{2}}$ vanish: the limit is $\frac{3}{1}=3$ on BOTH sides. Horizontal asymptote: $y=3$, the horizontal dashed line in the solution figure. A legitimate shortcut for a rational function with EQUAL degrees: the ratio of the leading coefficients, here $\frac{3}{1}$.

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b) With the product rule: $\begin{aligned}&h'(x) \\ &\quad = (2x)(x^{3}-2x)+(x^{2}+1)(3x^{2}-2) \\ &\quad = 2x^{4}-4x^{2}+3x^{4}-2x^{2}+3x^{2}-2 \\ &\quad = 5x^{4}-3x^{2}-2\end{aligned}$ Check by expanding first: $h(x)=x^{5}-x^{3}-2x$, so $h'=5x^{4}-3x^{2}-2$ ✓. Conclusion: both routes give the same result, but expanding first is only possible for polynomials; the product rule works on any product ($xe^{x}$, $x\sin x$). Pitfall: believing that $(fg)'=f'g'$. Here $f'g'=(2x)(3x^{2}-2)=6x^{3}-4x$, which has nothing to do with it.

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a) Three layers: the exponential, then the sine, then $2x$. $f'(x)=e^{\sin(2x)}\cdot\cos(2x)\cdot 2=2\cos(2x)e^{\sin(2x)}$. The factor 2 from the innermost layer is the one that gets forgotten: check at $x=0$, $f'(0)=2\cdot 1\cdot e^{0}=2$, whereas without the factor you would find 1, a slope half as large as it should be.

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c) $\frac{d}{dx}(xy)=1\cdot y+x\cdot y'=y+xy'$: this is the PRODUCT rule, and forgetting the term $y$ is the second most frequent mistake of the chapter. $\frac{d}{dx}(x^{2}y^{3})=2xy^{3}+x^{2}\cdot 3y^{2}y'=2xy^{3}+3x^{2}y^{2}y'$, product rule then chain rule. $\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{1\cdot y-x\cdot y'}{y^{2}}=\frac{y-xy'}{y^{2}}$, quotient rule.

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a) The large triangle, formed by the lamp and the tip of the shadow, and the small triangle, formed by the person and that same tip, are similar: same angles, two verticals and a common ground. The ratio of height to base is therefore the same: $\frac{6}{x+s}=\frac{1.8}{s}$. Cross-multiplying: $6s=1.8(x+s)$, that is $4.2s=1.8x$ and $s=\frac{3}{7}x$. The writing trap is here: the base of the LARGE triangle is $x+s$, not $x$. A solution that writes $\frac{6}{x}=\frac{1.8}{s}$ treats the shadow as if the lamp stood where the person is.

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a) $f(x)=\sqrt{x}$, $f'(x)=\frac{1}{2\sqrt{x}}$, so $f(4)=2$ and $f'(4)=\frac{1}{4}$. The linearization is $L(x)=2+\frac{1}{4}(x-4)$. Estimate: $\sqrt{4.1}\approx L(4.1)=2+\frac{0.1}{4}=2.025$. Exact value: $\sqrt{4.1}\approx 2.0248457$. The absolute error is about $0.000154$, a relative error of $0.0076$ percent. For mental arithmetic, that is remarkable: three correct decimals with no tool at all.

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c) The object speeds up (its SPEED, the absolute value of the velocity, increases) when $v$ and $a$ have the SAME sign, that is, when $v(t)\,a(t)>0$. Let us examine the four zones cut out by $t=2$, $t=3$, $t=4$. On $\left(0, 2\right)$: $v>0$ and $a<0$, a negative product, it SLOWS DOWN (it moves forward while braking towards its stop). On $\left(2, 3\right)$: $v<0$ and $a<0$, a positive product, it SPEEDS UP (it moves backward faster and faster). On $\left(3, 4\right)$: $v<0$ and $a>0$, a negative product, it SLOWS DOWN (it moves backward while braking). On $\left(4, +\infty\right)$: $v>0$ and $a>0$, a positive product, it SPEEDS UP. Conclusion: it speeds up on $\left(2, 3\right)\cup\left(4, +\infty\right)$. The key point is that a NEGATIVE acceleration does not mean slowing down: between $t=2$ and $t=3$, the acceleration is negative and yet the object goes faster and faster, because it is moving in the negative direction. It is the sign of the PRODUCT that decides, never the sign of $a$ alone.

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a) The displacement is $s(5)-s(0)=20-0=20$ metres. It is a difference of two positions, nothing more: the path followed in between plays no role. A displacement can even be zero although the particle has moved a lot, if it comes back to its starting point.

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a) If $f$ is continuous on $[a, b]$ and differentiable on $\left(a, b\right)$, then there is at least one $c$ in $\left(a, b\right)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Geometrically: there is at least one point of the curve where the TANGENT line is parallel to the SECANT line joining the two endpoints. When $f(a)=f(b)$, the secant line is horizontal and you get exactly Rolle's theorem.

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a) With $\sqrt{2}=1.41421356$: $e_{0}=\left|2-1.41421356\right|=0.58578644$; $e_{1}=\left|1.5-1.41421356\right|=0.08578644$; $e_{2}=\left|1.41666667-1.41421356\right|=0.00245310$; $e_{3}=\left|1.41421569-1.41421356\right|=2.1239\times 10^{-6}$.

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c) $\dfrac{x+1}{x^{2}+3}$ at $0$ equals $\dfrac{1}{3}$ by substitution: DETERMINATE. $\dfrac{\ln x}{x}$ at $0^{+}$ gives $\dfrac{-\infty}{0^{+}}$, which is not an indeterminate form: the limit is $-\infty$. The other two are indeterminate: $\dfrac{\infty}{\infty}$ for the third, $\dfrac{0}{0}$ for the fourth. Watch the trap in the second one: $\dfrac{0}{0}$ is indeterminate, but $\dfrac{\text{nonzero}}{0}$ never is.

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d) $h(x)=\frac{1}{x}=x^{-1}$ gives $h'(x)=-x^{-2}$ and $h''(x)=2x^{-3}=\frac{2}{x^{3}}$. For $x>0$: $h''>0$, the curve is concave up. For $x<0$: $h''<0$, it is concave down. The second derivative does change sign across $x=0$, but there is NO inflection point, because $x=0$ does not belong to the domain: there is no point of the curve with that $x$-coordinate. An inflection point is a POINT of the curve, so it needs a $y$-value. This is the classic pitfall of rational functions, and it is defused by always going back to the domain established in the first step of the analysis: hence the importance of the required order.

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a) Since the cylinder is inscribed and centred, its axis passes through the centre of the sphere and its two base circles lie at a distance $\frac{h}{2}$ from the centre. A radius of the sphere joining the centre to a point on the rim of a base forms a right triangle with legs $r$ and $\frac{h}{2}$ and hypotenuse $R$. The constraint is therefore $r^{2}+\left(\frac{h}{2}\right)^{2}=R^{2}$, here $r^{2}+\frac{h^{2}}{4}=9$.

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a) Differentiate both candidates. $F'(x)=2\tan x\cdot\sec^{2}x$ by the power rule and the chain rule, since $\left(\tan x\right)'=\sec^{2}x$. $G'(x)=2\sec x\cdot\left(\sec x\tan x\right)=2\sec^{2}x\tan x$, since $\left(\sec x\right)'=\sec x\tan x$. Both derivatives equal $f$, so both functions are indeed antiderivatives of $f$ ✓. Yet they do not look alike. The identity $\sec^{2}x-\tan^{2}x=1$ settles the matter: $G-F=1$, the constant that separates them is exactly $1$. Numerical check at $x=\frac{\pi}{4}$: $F=1$ and $G=2$ ✓. Two answers that look very different can therefore both be right, and this is the most common source of disputes over a test.

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