Solution. (a)(i). Write the root as a power and use the chain rule: \[y=\left (x^2+1\right )^{\frac {1}{2}},\qquad \frac {dy}{dx}=\tfrac {1}{2}\left (x^2+1\right )^{-\frac {1}{2}}\cdot 2x =x\left (x^2+1\right )^{-\frac {1}{2}}.\] Now the product rule, with a chain rule inside: \[\frac {d^2y}{dx^2}=1\cdot \left (x^2+1\right )^{-\frac {1}{2}} +x\cdot \left (-\tfrac {1}{2}\right )\left (x^2+1\right )^{-\frac {3}{2}}\cdot 2x =\left (x^2+1\right )^{-\frac {1}{2}}-x^2\left (x^2+1\right )^{-\frac {3}{2}}.\] Take out the lower power as a common factor: \[=\left (x^2+1\right )^{-\frac {3}{2}}\left [\left (x^2+1\right )-x^2\right ] =\frac {1}{\left (x^2+1\right )^{\frac {3}{2}}}.\]

(ii). Here \(\pi \) is just a number: \[\frac {dy}{dx}=3\pi x^2-7,\qquad \frac {d^2y}{dx^2}=6\pi x.\]

(iii). Write as a power: \[y=\left (x^2+1\right )^{-1},\qquad \frac {dy}{dx}=-\left (x^2+1\right )^{-2}\cdot 2x=\frac {-2x}{\left (x^2+1\right )^2}.\] Differentiating again by the product rule on \(-2x\left (x^2+1\right )^{-2}\): \[\frac {d^2y}{dx^2}=-2\left (x^2+1\right )^{-2} +(-2x)(-2)\left (x^2+1\right )^{-3}\cdot 2x =-2\left (x^2+1\right )^{-2}+8x^2\left (x^2+1\right )^{-3}\] \[=\left (x^2+1\right )^{-3}\left [-2\left (x^2+1\right )+8x^2\right ] =\frac {6x^2-2}{\left (x^2+1\right )^3}=\frac {2\left (3x^2-1\right )}{\left (x^2+1\right )^3}.\]

(iv). Quotient rule first: \[\frac {dy}{dx}=\frac {1(x-1)-(x+1)(1)}{(x-1)^2}=\frac {-2}{(x-1)^2}=-2(x-1)^{-2}.\] Then the chain rule on the power form: \[\frac {d^2y}{dx^2}=-2(-2)(x-1)^{-3}=\frac {4}{(x-1)^3}.\]

(b)(i). Write \(y=2\left (\sec x\right )^3\) and use the chain rule with \(\frac {d}{dx}\sec x=\sec x\tan x\): \[\frac {dy}{dx}=2\cdot 3\left (\sec x\right )^2\cdot \sec x\tan x =6\sec ^3x\tan x.\qquad \blacksquare \]

(b)(ii). With \(y=\left (x^2+1\right )^{\frac {1}{4}}\), \[\frac {dy}{dx}=\tfrac {1}{4}\left (x^2+1\right )^{-\frac {3}{4}}\cdot 2x =\tfrac {1}{2}x\left (x^2+1\right )^{-\frac {3}{4}}.\] Product rule again: \[\frac {d^2y}{dx^2}=\tfrac {1}{2}\left (x^2+1\right )^{-\frac {3}{4}} +\tfrac {1}{2}x\left (-\tfrac {3}{4}\right )\left (x^2+1\right )^{-\frac {7}{4}}\cdot 2x\] \[=\tfrac {1}{2}\left (x^2+1\right )^{-\frac {3}{4}} -\tfrac {3}{4}x^2\left (x^2+1\right )^{-\frac {7}{4}}.\] Factor out \(\frac {1}{4}\left (x^2+1\right )^{-\frac {7}{4}}\): \[=\tfrac {1}{4}\left (x^2+1\right )^{-\frac {7}{4}} \left [2\left (x^2+1\right )-3x^2\right ] =\frac {2-x^2}{4\left (x^2+1\right )^{\frac {7}{4}}}.\qquad \blacksquare \]

Solution. (a)(i). \[f'(x)=\tfrac {1}{2}\left (x^2+16\right )^{-\frac {1}{2}}\cdot 2x =\frac {x}{\sqrt {x^2+16}},\qquad f'(3)=\frac {3}{\sqrt {9+16}}=\frac {3}{5}.\]

(ii). By the quotient rule, \[f'(x)=\frac {1(x-4)-(x+3)(1)}{(x-4)^2}=\frac {-7}{(x-4)^2},\] \[f'(-3)=\frac {-7}{(-7)^2}=\frac {-7}{49}=-\frac {1}{7}.\]

(iii). Chain rule with \(\frac {d}{dx}\ln x=\frac {1}{x}\): \[f'(x)=2\ln x\cdot \frac {1}{x}=\frac {2\ln x}{x},\qquad f'(1)=\frac {2\ln 1}{1}=\frac {0}{1}=0.\]

(b)(i). \[f'(x)=3-3x^2=3\quad \implies \quad -3x^2=0\quad \implies \quad x=0.\] The \(y\)-coordinate is \(f(0)=5\), so the point is \((0,5)\).

(ii). \[f'(x)=\cos x=-1\quad \implies \quad x=\pi \ \text { (the only value in } [0,2\pi ]).\] Then \(f(\pi )=\sin \pi =0\), so the point is \((\pi ,0)\).

(iii). \[f'(x)=-\sin x=\tfrac {1}{2}\quad \implies \quad \sin x=-\tfrac {1}{2}.\] Sine is negative in quadrants III and IV, with reference angle \(\frac {\pi }{6}\): \[x=\tfrac {7\pi }{6}\quad \text {or}\quad x=\tfrac {11\pi }{6}.\] The corresponding \(y\)-values are \[\cos \tfrac {7\pi }{6}=-\tfrac {\sqrt {3}}{2},\qquad \cos \tfrac {11\pi }{6}=\tfrac {\sqrt {3}}{2},\] so the points are \(\left (\frac {7\pi }{6},-\frac {\sqrt {3}}{2}\right )\) and \(\left (\frac {11\pi }{6},\frac {\sqrt {3}}{2}\right )\).

Solution. Step 1 — confirm \(P\) is on the curve. \[y(0)=e^0\cos 0+1=1\cdot 1+1=2.\ \checkmark \]

Step 2 — differentiate. A product, with a chain rule in the first factor: \[\frac {dy}{dx}=\frac {d}{dx}\left (e^{2x}\right )\cos x +e^{2x}\frac {d}{dx}(\cos x)=2e^{2x}\cos x-e^{2x}\sin x =e^{2x}\left (2\cos x-\sin x\right ).\]

Step 3 — evaluate at \(x=0\). \[\left .\frac {dy}{dx}\right |_{x=0}=e^0\left (2\cos 0-\sin 0\right )=1(2-0)=2.\]

Step 4 — write down the two lines. Both pass through \((0,2)\).

Tangent, gradient \(2\): \[y-2=2(x-0)\quad \implies \quad y=2x+2.\]

Normal, gradient \(-\frac {1}{2}\): \[y-2=-\tfrac {1}{2}(x-0)\quad \implies \quad y=-\tfrac {1}{2}x+2, \quad \text {or}\quad x+2y=4.\]

Solution. (i). \(f'(x)=4x+3\), zero at \(x=-\frac {3}{4}\). It is negative to the left of that and positive to the right: \[\text {decreasing for } x<-\tfrac {3}{4},\qquad \text {increasing for } x>-\tfrac {3}{4}.\] An upward parabola with minimum at \(x=-\frac {3}{4}\).

(ii). \(f'(x)=-2x-4=-2(x+2)\), zero at \(x=-2\): \[\text {increasing for } x<-2,\qquad \text {decreasing for } x>-2.\] A downward parabola with maximum at \(x=-2\).

(iii). \(f'(x)=3x^2-2x-9\), which does not factorise. By the quadratic formula, \[x=\frac {2\pm \sqrt {4+108}}{6}=\frac {2\pm \sqrt {112}}{6}=\frac {1\pm 2\sqrt {7}}{3} \approx -1.43 \text { and } 2.10.\] Since \(f'\) is an upward parabola it is positive outside these values: \[\text {increasing for } x<\tfrac {1-2\sqrt {7}}{3} \text { and } x>\tfrac {1+2\sqrt {7}}{3},\qquad \text {decreasing between them}.\] For the sketch, note \(f(x)=x^2(x-1)-9(x-1)=(x-1)\left (x^2-9\right )=(x-1)(x-3)(x+3)\), so the roots are \(-3\), \(1\) and \(3\), and \(f(0)=9\).

(iv). Expand before differentiating: \[f(x)=2x\left (16-8x+x^2\right )=2x^3-16x^2+32x,\] \[f'(x)=6x^2-32x+32=2\left (3x^2-16x+16\right )=2(3x-4)(x-4),\] zero at \(x=\frac {4}{3}\) and \(x=4\). As an upward parabola \(f'\) is positive outside: \[\text {increasing for } x<\tfrac {4}{3} \text { and } x>4,\qquad \text {decreasing for } \tfrac {4}{3}<x<4.\]

(v). \(f'(x)=4x^3-2x=2x\left (2x^2-1\right )\), zero at \[x=0,\qquad x=\pm \tfrac {1}{\sqrt {2}}=\pm \tfrac {\sqrt {2}}{2}\approx \pm 0.707.\] Testing the sign in each of the four intervals:

A W-shape, with two minima at \(x=\pm \frac {\sqrt {2}}{2}\) and a local maximum at the origin.

Solution. (i). \(f'(x)=4x-4=0\) gives \(x=1\); \(f''(x)=4>0\), so a minimum, at \(f(1)=2-4+5=3\). Point \((1,3)\).

(ii). \(f'(x)=-4x-3=0\) gives \(x=-\frac {3}{4}\); \(f''(x)=-4<0\), so a maximum, at \[f\left (-\tfrac {3}{4}\right )=-2\left (\tfrac {9}{16}\right )+\tfrac {9}{4}+2 =-\tfrac {9}{8}+\tfrac {18}{8}+\tfrac {16}{8}=\tfrac {25}{8}.\] Point \(\left (-\frac {3}{4},\frac {25}{8}\right )\).

(iii). \(f'(x)=12x^2+18x-1=0\), which does not factorise: \[x=\frac {-18\pm \sqrt {324+48}}{24}=\frac {-18\pm \sqrt {372}}{24} =\frac {-9\pm \sqrt {93}}{12}\approx 0.054 \text { and } -1.554.\] \(f''(x)=24x+18\). At \(x=\frac {-9+\sqrt {93}}{12}\), \(f''=2\sqrt {93}>0\), a minimum; at \(x=\frac {-9-\sqrt {93}}{12}\), \(f''=-2\sqrt {93}<0\), a maximum. From Sheet 4 the roots are \(-2\), \(-1\) and \(\frac {3}{4}\), and \(f(0)=-6\).

(iv). \(f'(x)=3x^2-10x-22=0\): \[x=\frac {10\pm \sqrt {100+264}}{6}=\frac {10\pm \sqrt {364}}{6}=\frac {5\pm \sqrt {91}}{3} \approx -1.51 \text { and } 4.85.\] \(f''(x)=6x-10\). At \(x=\frac {5+\sqrt {91}}{3}\), \(f''=2\sqrt {91}>0\), a minimum; at \(x=\frac {5-\sqrt {91}}{3}\), \(f''=-2\sqrt {91}<0\), a maximum. From Sheet 4 the roots are \(-4\), \(2\) and \(7\), and \(f(0)=56\).

(v). \(f'(x)=3x^2-10x+3=(3x-1)(x-3)=0\), giving \(x=\frac {1}{3}\) and \(x=3\). \(f''(x)=6x-10\): \[f''\left (\tfrac {1}{3}\right )=2-10=-8<0\quad \implies \quad \textbf {maximum},\] \[f''(3)=18-10=8>0\quad \implies \quad \textbf {minimum}.\] The values are \[f\left (\tfrac {1}{3}\right )=\tfrac {1}{27}-\tfrac {5}{9}+1+9=\tfrac {256}{27} \approx 9.48,\qquad f(3)=27-45+9+9=0.\] Note \(f(3)=0\) is a repeated root — from Sheet 4, \(f(x)=(x-3)^2(x+1)\) — so the minimum sits exactly on the \(x\)-axis.

(vi). \(f'(x)=4x^3+12x^2-24x=4x\left (x^2+3x-6\right )=0\), giving \[x=0\qquad \text {and}\qquad x=\frac {-3\pm \sqrt {9+24}}{2}=\frac {-3\pm \sqrt {33}}{2} \approx 1.37 \text { and } -4.37.\] \(f''(x)=12x^2+24x-24=12\left (x^2+2x-2\right )\): \[f''(0)=-24<0\quad \implies \quad \textbf {maximum at } x=0,\ f(0)=0.\] For the other two, \(x^2=-3x+6\) from the first-derivative equation, so \[f''=12\left (-3x+6+2x-2\right )=12(4-x),\] giving \(f''\approx 12(2.63)>0\) and \(f''\approx 12(8.37)>0\): both minima. From Sheet 4, \(f(x)=x^2(x+6)(x-2)\), so the roots are \(-6\), \(0\) (twice) and \(2\).

Solution. (a). Check the point: \(\frac {(-1)^3}{5}=-\frac {1}{5}=-0.2\). \(\checkmark \) \[\frac {dy}{dx}=\frac {3x^2}{5},\qquad \left .\frac {dy}{dx}\right |_{x=-1}=\frac {3}{5}.\]

Tangent, gradient \(\frac {3}{5}\) through \((-1,-0.2)\): \[y+\tfrac {1}{5}=\tfrac {3}{5}(x+1)\quad \implies \quad y=\tfrac {3}{5}x+\tfrac {3}{5} -\tfrac {1}{5}=\tfrac {3}{5}x+\tfrac {2}{5}.\]

Normal, gradient \(-\frac {5}{3}\): \[y+\tfrac {1}{5}=-\tfrac {5}{3}(x+1)\quad \implies \quad y=-\tfrac {5}{3}x-\tfrac {5}{3}-\tfrac {1}{5}=-\tfrac {5}{3}x-\tfrac {28}{15}.\]

(b). \[P'(x)=-0.02x+550=0\quad \implies \quad x=\frac {550}{0.02}=27\,500.\] \[P''(x)=-0.02<0,\] so this is a maximum. The profit there is \[P(27\,500)=-0.01(27\,500)^2+550(27\,500)-10\,000\] \[=-7\,562\,500+15\,125\,000-10\,000=7\,552\,500.\] \[\therefore \quad \text {producing } 27\,500 \text { units gives a maximum profit of } 7\,552\,500.\]

(c). Step 1 — name the dimensions. The ends are square, so let each end be \(x\times x\) and let the length be \(y\).

Step 2 — use the volume to eliminate \(y\). \[x^2y=3.5\quad \implies \quad y=\frac {3.5}{x^2}.\]

Step 3 — write down the area. The box has two square ends, a base and two sides, but no lid: \[A=2x^2+3xy.\] Substituting for \(y\), \[A=2x^2+3x\cdot \frac {3.5}{x^2}=2x^2+\frac {10.5}{x}=2x^2+10.5x^{-1}.\]

Step 4 — minimise. \[\frac {dA}{dx}=4x-10.5x^{-2}=0\quad \implies \quad 4x=\frac {10.5}{x^2} \quad \implies \quad x^3=\frac {10.5}{4}=2.625,\] \[x=\sqrt [3]{2.625}\approx 1.3795.\]

Step 5 — confirm it is a minimum. \[\frac {d^2A}{dx^2}=4+21x^{-3},\] which is positive for every \(x>0\), so the stationary point is a minimum.

Step 6 — find the area. \[A=2(1.3795)^2+\frac {10.5}{1.3795}=3.807+7.611\approx 11.42.\] \[\therefore \quad \text {the least area of metal is about } 11.42\text { m}^2,\] with the box measuring about \(1.38\) m \(\times \ 1.38\) m \(\times \ 1.84\) m.

Solution. (a)(i). Write the first term as a power before differentiating: \[C(v)=1500v^{-1}+\tfrac {2}{11}v+60,\] \[\frac {dC}{dv}=-1500v^{-2}+\tfrac {2}{11}=0\quad \implies \quad \frac {1500}{v^2}=\frac {2}{11}.\] Cross-multiplying, \[2v^2=16\,500\quad \implies \quad v^2=8250\quad \implies \quad v=\sqrt {8250}=5\sqrt {330}\approx 90.83.\] Only the positive root is meaningful, since \(v\) is a speed. \[\therefore \quad v\approx 90.8\text { km/h}.\]

(ii). \[C(90.83)=\frac {1500}{90.83}+\frac {2(90.83)}{11}+60=16.51+16.51+60=93.03.\] \[\therefore \quad \text {the minimum cost is about K}93.03.\]

(b). Differentiating \(\frac {dC}{dv}=-1500v^{-2}+\frac {2}{11}\) again: \[\frac {d^2C}{dv^2}=3000v^{-3}=\frac {3000}{v^3}.\] For any speed \(v>0\) this is positive, and in particular \[\left .\frac {d^2C}{dv^2}\right |_{v=90.83}=\frac {3000}{(90.83)^3}\approx 0.004>0,\] so the stationary point is a minimum. \(\blacksquare \)

Solution. The limit exists exactly when the two one-sided limits agree.

Two worth setting out.

(e). From the left the first rule applies, from the right the second: \[\lim _{x\to 1^-}(x-1)=0,\qquad \lim _{x\to 1^+}(-x+1)=0.\] They agree, so \(\lim _{x\to 1}f(x)=0\). The two pieces meet, and the graph has a corner there but no break.

(f). Substituting gives \(\frac {0}{0}\), so factorise: \[\frac {x^2-x-2}{x-2}=\frac {(x-2)(x+1)}{x-2}=x+1\quad (x\neq 2),\] \[\lim _{x\to 2}f(x)=\lim _{x\to 2}(x+1)=3.\] Note \(f(2)\) itself does not exist, yet the limit does.

Solution.

(a). At \(x=5\) the numerator is \(25-5-5=15\) and the denominator is \(0\). This is \(\frac {15}{0}\), not \(\frac {0}{0}\): \[\lim _{x\to 5^-}\frac {x^2-x-5}{x-5}=-\infty ,\qquad \lim _{x\to 5^+}\frac {x^2-x-5}{x-5}=+\infty ,\] since the numerator stays near \(15\) while the denominator is small and negative on the left, small and positive on the right. \[\therefore \quad \text {the limit \textbf {does not exist}}.\]

(b). At \(x=3\) the numerator is \(6\) and the denominator \(0\). Same situation: \[\therefore \quad \text {the limit \textbf {does not exist}} \text { (}-\infty \text { from the left, } +\infty \text { from the right)}.\]

(c). At \(x=3\) the numerator is \(9-3=6\) and the denominator \(9-9=0\). Again \(\frac {6}{0}\): \[\therefore \quad \text {the limit \textbf {does not exist}}.\]

(d). Substituting gives \(\frac {2(9)-18}{0}=\frac {0}{0}\), so expand: \[2(-3+x)^2-18=2\left (9-6x+x^2\right )-18=18-12x+2x^2-18=2x^2-12x=x(2x-12).\] \[\therefore \quad \lim _{x\to 0}\frac {x(2x-12)}{x}=\lim _{x\to 0}(2x-12)=-12.\]

(e). Inside the root, substituting gives \(\frac {4-6+2}{4-4}=\frac {0}{0}\), so factorise: \[\frac {x^2-3x+2}{x^2-4}=\frac {(x-1)(x-2)}{(x-2)(x+2)}=\frac {x-1}{x+2}\quad (x\neq 2).\] \[\therefore \quad \lim _{x\to 2}\sqrt {\frac {x-1}{x+2}}=\sqrt {\frac {1}{4}}=\frac {1}{2}.\]

(f). \(\frac {0}{0}\), and \(x^3-1\) is a difference of cubes: \[\frac {x^3-1}{x-1}=\frac {(x-1)\left (x^2+x+1\right )}{x-1}=x^2+x+1,\] \[\therefore \quad \lim _{x\to 1}\left (x^2+x+1\right )=1+1+1=3.\]

Solution. For a polynomial, the highest power decides everything. For a rational function, divide top and bottom by the highest power in the denominator and use \(\frac {1}{x^n}\to 0\).

(a). The leading term is \(3x^4\). As \(x\to -\infty \), \(x^4\) is positive and grows without bound: \[\lim _{x\to -\infty }\left (3x^4-x^2+x-7\right )=+\infty .\]

(b). Divide by \(x^2\): \[\frac {\frac {2}{x}+\frac {5}{x^2}}{1-\frac {7}{x}+\frac {3}{x^2}} \longrightarrow \frac {0+0}{1-0+0}=0.\]

(c). Divide by \(x^3\): \[\frac {3-\frac {4}{x^2}+\frac {2}{x^3}}{7+\frac {5}{x^3}}\longrightarrow \frac {3}{7}.\]

(d). Divide by \(x\): \[\frac {2-\frac {1}{x}}{1-\frac {1}{x}}\longrightarrow \frac {2}{1}=2.\]

(e). Divide by \(x^3\): \[\frac {4x^2-\frac {1}{x^3}}{3+\frac {7}{x^3}}\longrightarrow \frac {4x^2}{3},\] and \(4x^2\) grows without bound, so the limit is \(+\infty \) — it does not exist as a finite number.

(f). The leading term is \(2x^{11}\), which grows without bound as \(x\to \infty \): \[\lim _{x\to \infty }\left (2x^{11}-5x^6+3x^2+1\right )=+\infty .\]

Solution. Recall the three conditions: \(f(a)\) defined, the limit existing, and the two being equal.

(a). \(f(0)=0^2=0\), defined. The one-sided limits are \[\lim _{x\to 0^-}x=0,\qquad \lim _{x\to 0^+}x^2=0,\] which agree, so the limit is \(0\) and equals \(f(0)\). All three conditions hold: \(f\) is continuous at \(x=0\).

(b). \(f(0)=1\), defined. But \[\lim _{x\to 0^-}(-1)=-1,\qquad \lim _{x\to 0^+}1=1,\] which disagree, so the limit does not exist. The second condition fails: \(f\) is discontinuous at \(x=0\), with a jump of size \(2\).

(c). At \(x=-2\) the denominator is zero, so \(f(-2)\) is not defined and the first condition fails outright: \(f\) is discontinuous at \(x=-2\). The limit nevertheless exists, \[\lim _{x\to -2}\frac {(x-2)(x+2)}{x+2}=\lim _{x\to -2}(x-2)=-4,\] so the graph has a hole at \((-2,-4)\) rather than a break. Defining \(f(-2)=-4\) would repair it.

(d). At \(x=1\): \(f(1)=1-1=0\), defined. The one-sided limits are \[\lim _{x\to 1^-}(x-1)=0,\qquad \lim _{x\to 1^+}(2x-1)=1,\] which disagree, so \(f\) is discontinuous at \(x=1\).

At \(x=2\): the definition covers \(x\leq 1\) and \(1<x<2\) but says nothing about \(x\geq 2\), so \(f(2)\) is not defined and \(f\) is discontinuous at \(x=2\) on the first condition.

Solution. Throughout, \[f'(x)=\lim _{h\to 0}\frac {f(x+h)-f(x)}{h}.\]

(a). \[\frac {7(x+h)-7x}{h}=\frac {7h}{h}=7\quad \implies \quad f'(x)=7.\]

(b). \[\frac {3(x+h)^2+5-\left (3x^2+5\right )}{h}=\frac {6xh+3h^2}{h}=6x+3h \quad \implies \quad f'(x)=6x.\]

(c). Combine over a common denominator first: \[\frac {1}{(x+h)^2+2}-\frac {1}{x^2+2} =\frac {\left (x^2+2\right )-\left ((x+h)^2+2\right )}{\left [(x+h)^2+2\right ]\left (x^2+2\right )} =\frac {-2xh-h^2}{\left [(x+h)^2+2\right ]\left (x^2+2\right )}.\] Dividing by \(h\) cancels one factor of \(h\): \[\frac {-2x-h}{\left [(x+h)^2+2\right ]\left (x^2+2\right )} \longrightarrow \frac {-2x}{\left (x^2+2\right )^2}.\]

(d). A difference of surds, so rationalise the numerator: \[\frac {\sqrt {x+h-1}-\sqrt {x-1}}{h} \times \frac {\sqrt {x+h-1}+\sqrt {x-1}}{\sqrt {x+h-1}+\sqrt {x-1}} =\frac {(x+h-1)-(x-1)}{h\left (\sqrt {x+h-1}+\sqrt {x-1}\right )}\] \[=\frac {h}{h\left (\sqrt {x+h-1}+\sqrt {x-1}\right )} =\frac {1}{\sqrt {x+h-1}+\sqrt {x-1}}\longrightarrow \frac {1}{2\sqrt {x-1}}.\]

(e). Combine, then rationalise: \[\frac {1}{\sqrt {2x+2h+1}}-\frac {1}{\sqrt {2x+1}} =\frac {\sqrt {2x+1}-\sqrt {2x+2h+1}}{\sqrt {2x+2h+1}\sqrt {2x+1}}.\] Multiplying top and bottom by \(\sqrt {2x+1}+\sqrt {2x+2h+1}\) turns the numerator into \((2x+1)-(2x+2h+1)=-2h\), so after dividing by \(h\), \[\frac {-2}{\sqrt {2x+2h+1}\sqrt {2x+1}\left (\sqrt {2x+1}+\sqrt {2x+2h+1}\right )} \longrightarrow \frac {-2}{(2x+1)\cdot 2\sqrt {2x+1}}=-\frac {1}{(2x+1)^{\frac {3}{2}}}.\]

(f). Rationalise as in (d): \[\frac {\sqrt {2x+2h}-\sqrt {2x}}{h}=\frac {2h}{h\left (\sqrt {2x+2h}+\sqrt {2x}\right )} =\frac {2}{\sqrt {2x+2h}+\sqrt {2x}}\longrightarrow \frac {2}{2\sqrt {2x}} =\frac {1}{\sqrt {2x}}=\frac {\sqrt {2}}{2\sqrt {x}}.\]

Solution. Each becomes a sum of powers once rewritten, after which the power rule is enough.

(a). \(y=x+x^{-1}\), so \[\frac {dy}{dx}=1-x^{-2}=1-\frac {1}{x^2}.\]

(b). \(\frac {8}{\sqrt {x}}=8x^{-\frac {1}{2}}\), so \(y=x^2-8x^{-\frac {1}{2}}\) and \[\frac {dy}{dx}=2x-8\left (-\tfrac {1}{2}\right )x^{-\frac {3}{2}} =2x+4x^{-\frac {3}{2}}=2x+\frac {4}{x^{\frac {3}{2}}}.\]

(c). Expand: \[y=x^2+4x+4\quad \implies \quad \frac {dy}{dx}=2x+4.\]

(d). Divide each term by \(x^4\): \[y=3x^3+5x-2+x^{-3}-3x^{-4},\] \[\frac {dy}{dx}=9x^2+5-3x^{-4}+12x^{-5}=9x^2+5-\frac {3}{x^4}+\frac {12}{x^5}.\]

Solution. (a). Chain rule, with the inner derivative \(3x^2-4-\frac {1}{x^2}\): \[f'(x)=4\left (x^3-4x+\tfrac {1}{x}\right )^3\left (3x^2-4-\tfrac {1}{x^2}\right ).\]

(b). Chain rule twice: \[f'(x)=3\sin ^2(2x+5)\cdot \cos (2x+5)\cdot 2=6\sin ^2(2x+5)\cos (2x+5).\]

(c). Quotient rule: \[s'(t)=\frac {2(2t-1)-(2t+1)(2)}{(2t-1)^2}=\frac {4t-2-4t-2}{(2t-1)^2} =\frac {-4}{(2t-1)^2}.\]

(d). Write \(\sqrt {\sin x}=(\sin x)^{\frac {1}{2}}\): \[h'(x)=\cos x+\tfrac {1}{2}(\sin x)^{-\frac {1}{2}}\cos x =\cos x\left (1+\frac {1}{2\sqrt {\sin x}}\right ).\]

(e). Product rule, with a chain rule on the first factor: \[f'(x)=5(4x+2)^4\cdot 4\cdot \left (3x^2-x\right )+(4x+2)^5(6x-1)\] \[=(4x+2)^4\left [20\left (3x^2-x\right )+(4x+2)(6x-1)\right ] =(4x+2)^4\left (84x^2-14x-2\right ).\]

(f). Product rule with a chain rule on \(e^{-2x}\): \[f'(x)=2xe^{-2x}+x^2\left (-2e^{-2x}\right )=2xe^{-2x}(1-x).\]

(g). Chain rule with \(\frac {d}{dx}\ln u=\frac {u'}{u}\): \[f'(x)=\frac {4\cos 4x}{\sin 4x}=4\cot 4x.\]

(h). Simplify the numerator before differentiating — this is the whole question: \[\ln \left (e^x\cdot 2x\right )=\ln e^x+\ln 2x=x+\ln 2+\ln x.\] So with \(u=x+\ln 2+\ln x\) and \(v=\ln (x+1)\), giving \(u'=1+\frac {1}{x}\) and \(v'=\frac {1}{x+1}\), \[f'(x)=\frac {\left (1+\frac {1}{x}\right )\ln (x+1) -\left (x+\ln 2+\ln x\right )\cdot \frac {1}{x+1}}{\left [\ln (x+1)\right ]^2}.\]

(i). Use the double-angle identity first: \[\sin A\cos A=\tfrac {1}{2}\sin 2A\quad \implies \quad h(x)=\tfrac {1}{2}\sin (4x+2),\] \[h'(x)=\tfrac {1}{2}\cos (4x+2)\cdot 4=2\cos (4x+2).\]

(j). Quotient rule, with a chain rule on \(\sin x^2\): \[k'(x)=\frac {2x\cos \left (x^2\right )\cdot x^2-\sin \left (x^2\right )\cdot 2x}{x^4} =\frac {2x^2\cos \left (x^2\right )-2\sin \left (x^2\right )}{x^3}.\]

Solution. Differentiate every term with respect to \(x\), remembering that each \(y\) contributes a factor \(\frac {dy}{dx}\), then collect and factor.

(a). Note \(\ln xy=\ln x+\ln y\), which avoids a product rule: \[3\frac {dy}{dx}-2x+\frac {1}{x}+\frac {1}{y}\frac {dy}{dx}=0,\] \[\frac {dy}{dx}\left (3+\frac {1}{y}\right )=2x-\frac {1}{x}\quad \implies \quad \frac {dy}{dx}=\frac {2x-\frac {1}{x}}{3+\frac {1}{y}} =\frac {y\left (2x^2-1\right )}{x(3y+1)}.\]

(b). The middle term needs the product rule: \[3y^2\frac {dy}{dx}+2x\ln y+x^2\cdot \frac {1}{y}\frac {dy}{dx}=5,\] \[\frac {dy}{dx}\left (3y^2+\frac {x^2}{y}\right )=5-2x\ln y\quad \implies \quad \frac {dy}{dx}=\frac {y\left (5-2x\ln y\right )}{3y^3+x^2}.\]

(c). Both terms are products: \[\ln y+\frac {x}{y}\frac {dy}{dx}-\left (\frac {dy}{dx}\ln x+\frac {y}{x}\right )=0,\] \[\frac {dy}{dx}\left (\frac {x}{y}-\ln x\right )=\frac {y}{x}-\ln y \quad \implies \quad \frac {dy}{dx}=\frac {y\left (y-x\ln y\right )}{x\left (x-y\ln x\right )}.\]

(d). The exponent \(xy\) needs the product rule inside the chain rule: \[e^{xy}\left (y+x\frac {dy}{dx}\right )-3x^2+6y\frac {dy}{dx}=0,\] \[\frac {dy}{dx}\left (xe^{xy}+6y\right )=3x^2-ye^{xy}\quad \implies \quad \frac {dy}{dx}=\frac {3x^2-ye^{xy}}{xe^{xy}+6y}.\]

(e). \[e^y+xe^y\frac {dy}{dx}+2-\frac {1}{y+1}\frac {dy}{dx}=0,\] \[\frac {dy}{dx}\left (xe^y-\frac {1}{y+1}\right )=-e^y-2\quad \implies \quad \frac {dy}{dx}=\frac {-\left (e^y+2\right )(y+1)}{x e^y(y+1)-1}.\]

(f). \[2+6y\frac {dy}{dx}+6xy+3x^2\frac {dy}{dx}=0,\] \[\frac {dy}{dx}\left (6y+3x^2\right )=-2-6xy\quad \implies \quad \frac {dy}{dx}=\frac {-(2+6xy)}{3x^2+6y}.\]

Solution. (a). Differentiating, \[2x-y-x\frac {dy}{dx}+2y\frac {dy}{dx}=0\quad \implies \quad \frac {dy}{dx}=\frac {y-2x}{2y-x}.\] At \((1,2)\): \(\frac {2-2}{4-1}=\frac {0}{3}=0\). The tangent is horizontal there.

(c). Differentiating \(x^2+y^2=25\) gives \[2x+2y\frac {dy}{dx}=0\quad \implies \quad \frac {dy}{dx}=-\frac {x}{y}.\] At \((5,0)\) this is \(-\frac {5}{0}\), which is undefined. The point \((5,0)\) is the right-hand end of the circle of radius \(5\), where the tangent is vertical — and a vertical line has no gradient. The answer is that \(\frac {dy}{dx}\) does not exist there, not that it is zero or infinite.

(e). Differentiating, \[4x+6y\frac {dy}{dx}-1+6y+6x\frac {dy}{dx}=0\quad \implies \quad \frac {dy}{dx}=\frac {1-4x-6y}{6y+6x}.\] At \((1,-2)\): \[\frac {1-4+12}{-12+6}=\frac {9}{-6}=-\frac {3}{2}.\]

(f). Differentiating, \[3(x+y)^2\left (1+\frac {dy}{dx}\right )=2x+\frac {dy}{dx},\] \[\frac {dy}{dx}\left [3(x+y)^2-1\right ]=2x-3(x+y)^2\quad \implies \quad \frac {dy}{dx}=\frac {2x-3(x+y)^2}{3(x+y)^2-1}.\] At \((1,0)\), \((x+y)^2=1\): \[\frac {2-3}{3-1}=-\frac {1}{2}.\]

(b). From \(x^3y^2=2\), at \(x=-5\) we need \(y\) first: \[-125y^2=2\quad \implies \quad y^2=-\tfrac {2}{125},\] which has no real solution. There is no point on the curve with \(x=-5\), so the question has no answer over the real numbers.

(d). At \(x=2\), \(4+6y^2-y^3=9\), that is \(y^3-6y^2+5=0\). Testing the divisors of \(5\) gives \(y=1\): \(1-6+5=0\). Dividing, \[y^3-6y^2+5=(y-1)\left (y^2-5y-5\right ),\] so \(y=1\) or \(y=\frac {5\pm \sqrt {45}}{2}\). Taking the convenient point \((2,1)\) and differentiating the original, \[2x+3y^2+6xy\frac {dy}{dx}-3y^2\frac {dy}{dx}=0\quad \implies \quad \frac {dy}{dx}=\frac {-2x-3y^2}{6xy-3y^2}.\] At \((2,1)\): \[\frac {-4-3}{12-3}=-\frac {7}{9}.\]

Solution. Find the stationary points inside the interval, then compare the function’s value there with its value at both endpoints.

(a). \(y'=4x-4=0\) gives \(x=1\), inside \([0,3]\). \[y(0)=2,\qquad y(1)=0,\qquad y(3)=18-12+2=8.\] \[\therefore \quad \text {maximum } 8 \text { at } x=3,\qquad \text {minimum } 0 \text { at } x=1.\]

(b). \(y'=3x^2-3=0\) gives \(x=\pm 1\), but only \(x=1\) lies in \([0,2]\). \[y(0)=1,\qquad y(1)=1-3+1=-1,\qquad y(2)=8-6+1=3.\] \[\therefore \quad \text {maximum } 3 \text { at } x=2,\qquad \text {minimum } -1 \text { at } x=1.\]

(c). \(y'=3x^2+6x-9=3(x+3)(x-1)=0\) gives \(x=-3\) and \(x=1\), both inside \([-4,4]\). \[y(-4)=-64+48+36+2=22,\qquad y(-3)=-27+27+27+2=29,\] \[y(1)=1+3-9+2=-3,\qquad y(4)=64+48-36+2=78.\] \[\therefore \quad \text {maximum } 78 \text { at } x=4,\qquad \text {minimum } -3 \text { at } x=1.\]

(d). By the quotient rule, \[y'=\frac {-\left (x^2+3x\right )-(1-x)(2x+3)}{\left (x^2+3x\right )^2} =\frac {x^2-2x-3}{\left (x^2+3x\right )^2}=\frac {(x-3)(x+1)}{\left (x^2+3x\right )^2}.\] The numerator is zero at \(x=3\) and \(x=-1\); only \(x=3\) lies in \([1,4]\). \[y(1)=\frac {0}{4}=0,\qquad y(3)=\frac {-2}{18}=-\frac {1}{9},\qquad y(4)=\frac {-3}{28}.\] Since \(-\frac {1}{9}\approx -0.111\) and \(-\frac {3}{28}\approx -0.107\), \[\therefore \quad \text {maximum } 0 \text { at } x=1,\qquad \text {minimum } -\tfrac {1}{9} \text { at } x=3.\]

Solution. (a). The marginal cost is \(C'(x)\). Differentiate the average cost by the quotient rule: \[c(x)=\frac {C(x)}{x}\quad \implies \quad c'(x)=\frac {C'(x)\cdot x-C(x)\cdot 1}{x^2}.\] At a minimum of the average cost, \(c'(x)=0\), so the numerator is zero: \[xC'(x)-C(x)=0\quad \implies \quad C'(x)=\frac {C(x)}{x}=c(x).\] That is, marginal cost equals average cost. \(\blacksquare \)

(b)(i). With \(C(x)=16\,000+200x+4x^{3/2}\) and \(x=1000\), using \(1000^{3/2}=1000\sqrt {1000}\approx 31\,622.78\): \[C(1000)=16\,000+200\,000+4(31\,622.78)\approx 342\,491.1,\] \[c(1000)=\frac {342\,491.1}{1000}\approx 342.49\ \text {per unit}.\] The marginal cost is \[C'(x)=200+6x^{\frac {1}{2}}\quad \implies \quad C'(1000)=200+6\sqrt {1000} \approx 200+189.74=389.74.\]

(ii). The average cost is \[c(x)=\frac {16\,000}{x}+200+4x^{\frac {1}{2}}=16\,000x^{-1}+200+4x^{\frac {1}{2}}.\] \[c'(x)=-16\,000x^{-2}+2x^{-\frac {1}{2}}=0\quad \implies \quad \frac {2}{\sqrt {x}}=\frac {16\,000}{x^2}\quad \implies \quad 2x^2=16\,000\sqrt {x}.\] Dividing by \(2\sqrt {x}\), \[x^{\frac {3}{2}}=8000\quad \implies \quad x=8000^{\frac {2}{3}}=\left (20^3\right )^{2/3} =400.\]

(iii). \[c(400)=\frac {16\,000}{400}+200+4\sqrt {400}=40+200+80=320.\] \[\therefore \quad \text {the minimum average cost is } 320 \text { per unit, at } 400 \text { units}.\]

Check against part (a). The marginal cost at \(x=400\) is \[C'(400)=200+6\sqrt {400}=200+120=320,\] which equals the average cost of \(320\), exactly as part (a) predicts. \(\checkmark \)

Solution. Step 1 — build the revenue. The demand function gives the price at which \(x\) units sell, so \[R(x)=x\,p(x)=x(1700-7x)=1700x-7x^2.\]

Step 2 — build the profit. \[P(x)=R(x)-C(x)=\left (1700x-7x^2\right ) -\left (16\,000+500x-1.6x^2+0.004x^3\right )\] \[=-0.004x^3-5.4x^2+1200x-16\,000.\]

Step 3 — differentiate and solve. \[P'(x)=-0.012x^2-10.8x+1200=0.\] Dividing through by \(-0.012\): \[x^2+900x-100\,000=0\quad \implies \quad (x+1000)(x-100)=0,\] \[x=100\quad \text {or}\quad x=-1000.\] A negative output is meaningless, so \(x=100\).

Step 4 — confirm it is a maximum. \[P''(x)=-0.024x-10.8,\qquad P''(100)=-2.4-10.8=-13.2<0,\] so the profit is greatest at \(x=100\).

Step 5 — the profit. \[P(100)=-0.004(10^6)-5.4(10^4)+1200(100)-16\,000\] \[=-4000-54\,000+120\,000-16\,000=46\,000.\]

\[\therefore \quad \text {producing } 100 \text { units maximises profit, at } 46\,000.\]