Solution. (a). By definition, \begin {align*} f'(x)&=\lim _{h\to 0}\frac {\sqrt {x+h+1}-\sqrt {x+1}}{h}\\ &=\lim _{h\to 0}\frac {\left (\sqrt {x+h+1}-\sqrt {x+1}\right ) \left (\sqrt {x+h+1}+\sqrt {x+1}\right )} {h\left (\sqrt {x+h+1}+\sqrt {x+1}\right )}\\ &=\lim _{h\to 0}\frac {(x+h+1)-(x+1)}{h\left (\sqrt {x+h+1}+\sqrt {x+1}\right )} =\lim _{h\to 0}\frac {1}{\sqrt {x+h+1}+\sqrt {x+1}}\\ &=\frac {1}{2\sqrt {x+1}} . \end {align*}

(b). Quotient rule with \(u=\left (2x^{2}-3x+5\right )^{3}\) and \(v=x-2\). The numerator needs the chain rule: \[\frac {du}{dx}=3\left (2x^{2}-3x+5\right )^{2}(4x-3),\quad \frac {dv}{dx}=1 .\] Hence \[\frac {dy}{dx} =\frac {3(x-2)\left (2x^{2}-3x+5\right )^{2}(4x-3)-\left (2x^{2}-3x+5\right )^{3}} {(x-2)^{2}} .\] Taking out the common factor \(\left (2x^{2}-3x+5\right )^{2}\), \[\frac {dy}{dx}=\frac {\left (2x^{2}-3x+5\right )^{2} \left [3(x-2)(4x-3)-\left (2x^{2}-3x+5\right )\right ]}{(x-2)^{2}} =\frac {\left (2x^{2}-3x+5\right )^{2}\left (10x^{2}-30x+13\right )}{(x-2)^{2}} .\]

Solution. (a). Product rule with \(u=x\) and \(v=\ln \left (x^{2}+3x\right )\), the second needing the chain rule: \[\frac {dy}{dx}=\ln \left (x^{2}+3x\right )+x\cdot \frac {2x+3}{x^{2}+3x} =\ln \left (x^{2}+3x\right )+\frac {2x+3}{x+3},\] cancelling one factor of \(x\) in the second term.

(b). Simplify before differentiating, which is far quicker than the quotient rule here: \[y=\frac {\sec ^{2}x}{\tan x} =\frac {1}{\cos ^{2}x}\cdot \frac {\cos x}{\sin x} =\frac {1}{\sin x\cos x}=\frac {2}{\sin 2x}=2\operatorname {cosec}2x .\] Then \[\frac {dy}{dx}=-2\operatorname {cosec}2x\cot 2x\cdot 2 =-4\operatorname {cosec}2x\cot 2x .\]

Solution. First confirm the point is on the curve: \(16+16+40+16-88=0\), so it is.

Differentiating implicitly, \[2x+2y\frac {dy}{dx}-10+4\frac {dy}{dx}=0 \implies \frac {dy}{dx}=\frac {10-2x}{2y+4}=\frac {5-x}{y+2}.\] At \((-4,4)\) this is \(\frac {5+4}{4+2}=\frac {9}{6}=\frac {3}{2}\), the gradient of the tangent. The normal is perpendicular, so its gradient is \(-\frac {2}{3}\), and \[y-4=-\frac {2}{3}(x+4) \implies y=-\frac {2}{3}x+\frac {20}{3}.\]

Solution. (a). The rectangle has width \(2r\) — the diameter of the semicircle — and height \(x\), say. The perimeter consists of the two vertical sides, the base, and the semicircular arc: \[2x+2r+\pi r=40 \implies x=20-r-\frac {\pi r}{2}.\] The area is the rectangle plus the semicircle: \begin {align*} A&=2rx+\frac {1}{2}\pi r^{2} =2r\left (20-r-\frac {\pi r}{2}\right )+\frac {1}{2}\pi r^{2}\\ &=40r-2r^{2}-\pi r^{2}+\frac {1}{2}\pi r^{2} =40r-2r^{2}-\frac {\pi r^{2}}{2}, \end {align*}

as required.

(b). Differentiate and set to zero: \[\frac {dA}{dr}=40-4r-\pi r=0 \implies r(4+\pi )=40 \implies r=\frac {40}{4+\pi } \approx 5.60\ \text {cm}.\] This is a maximum, since \(\frac {d^{2}A}{dr^{2}}=-4-\pi <0\) everywhere.

Solution. (a). With height \(h\), the volume is \(\pi x^{2}h=1000\), so \(h=\frac {1000}{\pi x^{2}}\). The total surface is two circular ends plus the curved side: \[A=2\pi x^{2}+2\pi x h=2\pi x^{2}+2\pi x\cdot \frac {1000}{\pi x^{2}} =2\pi x^{2}+\frac {2000}{x}.\]

(b). Differentiate, writing \(\frac {2000}{x}\) as \(2000x^{-1}\): \[\frac {dA}{dx}=4\pi x-\frac {2000}{x^{2}} .\] Setting this to zero, \(4\pi x^{3}=2000\), so \[x^{3}=\frac {2000}{4\pi }=\frac {500}{\pi }.\] This is a minimum, since \[\frac {d^{2}A}{dx^{2}}=4\pi +\frac {4000}{x^{3}}>0\] for every \(x>0\).