Companion to the University-Style Practice Set大学风格练习题的配套解答
Sections 1 to 7: power/sum/constant-multiple rules, product rule, quotient rule, chain rule, combining rules, polynomials and rational functions, pitfalls第 1 至 7 节：幂法则/求和法则/常数倍法则、乘积法则、商法则、链式法则、综合运用、多项式与有理函数、常见错误CALC I
Differentiate: (a) $f(x)=4x^{5}-\frac{3}{2}x^{4}+7x-9$; (b) $g(x)=\frac{5}{x^{3}}-\frac{2}{\sqrt{x}}+6x^{2/3}$; (c) $h(x)=\pi x^{4}-e\,x^{2}+\sqrt{3}$.
Apply the constant-multiple and power rules to each term, and the sum rule to add them. The constant $-9$ differentiates to $0$. (M1)对每一项应用常数倍法则和幂法则，用求和法则将各项相加。常数项 $-9$ 的导数为 $0$。(M1)
$$ f'(x)=4\cdot 5x^{4}-\tfrac{3}{2}\cdot 4x^{3}+7\cdot 1-0=20x^{4}-6x^{3}+7. $$(A1)
Rewrite as $g(x)=5x^{-3}-2x^{-1/2}+6x^{2/3}$ and apply the power rule to each term. (M1)将其改写为 $g(x)=5x^{-3}-2x^{-1/2}+6x^{2/3}$，再对每项应用幂法则。(M1)
$$ g'(x)=5\cdot(-3)x^{-4}-2\cdot\left(-\tfrac{1}{2}\right)x^{-3/2}+6\cdot\tfrac{2}{3}x^{-1/3}=-15x^{-4}+x^{-3/2}+4x^{-1/3}. $$(A1)
$\pi$ and $e$ are constants, so the constant-multiple rule applies exactly as for any real number coefficient. The term $\sqrt{3}$ is a constant, so its derivative is $0$. (M1)$\pi$ 与 $e$ 均为常数，因此常数倍法则的应用方式与任何实数系数完全相同。$\sqrt{3}$ 也是常数，其导数为 $0$。(M1)
$$ h'(x)=\pi\cdot 4x^{3}-e\cdot 2x+0=4\pi x^{3}-2ex. $$(A1)
Differentiate without expanding first: (a) $f(x)=(3x^{2}+1)(x^{3}-2x)$; (b) $g(x)=x^{4}(2x^{2}-5)^{3}$; (c) $h(x)=(x^{2}+3)(x^{2}-3)$ by the product rule and then by expanding; verify they agree.
Let $u=3x^{2}+1$ and $v=x^{3}-2x$, so $u'=6x$ and $v'=3x^{2}-2$. The product rule gives $(uv)'=u'v+uv'$. (M1)令 $u=3x^{2}+1$，$v=x^{3}-2x$，则 $u'=6x$，$v'=3x^{2}-2$。乘积法则给出 $(uv)'=u'v+uv'$。(M1)
$$ f'(x)=6x\left(x^{3}-2x\right)+\left(3x^{2}+1\right)\!\left(3x^{2}-2\right). $$Expand term by term: $6x^{4}-12x^{2}$ and $9x^{4}-6x^{2}+3x^{2}-2=9x^{4}-3x^{2}-2$. (A1)逐项展开：$6x^{4}-12x^{2}$ 与 $9x^{4}-6x^{2}+3x^{2}-2=9x^{4}-3x^{2}-2$。(A1)
$$ f'(x)=6x^{4}-12x^{2}+9x^{4}-3x^{2}-2=15x^{4}-15x^{2}-2. $$(A1)
Let $u=x^{4}$ (so $u'=4x^{3}$) and $v=(2x^{2}-5)^{3}$ (chain rule: $v'=3(2x^{2}-5)^{2}\cdot 4x=12x(2x^{2}-5)^{2}$). (M1)令 $u=x^{4}$（则 $u'=4x^{3}$），$v=(2x^{2}-5)^{3}$（链式法则：$v'=3(2x^{2}-5)^{2}\cdot 4x=12x(2x^{2}-5)^{2}$）。(M1)
$$ g'(x)=4x^{3}\!\left(2x^{2}-5\right)^{3}+x^{4}\cdot 12x\!\left(2x^{2}-5\right)^{2}=4x^{3}\!\left(2x^{2}-5\right)^{3}+12x^{5}\!\left(2x^{2}-5\right)^{2}. $$Factor out $4x^{3}(2x^{2}-5)^{2}$: (A1)提取公因子 $4x^{3}(2x^{2}-5)^{2}$：(A1)
$$ g'(x)=4x^{3}\!\left(2x^{2}-5\right)^{2}\!\left[\left(2x^{2}-5\right)+3x^{2}\right]=4x^{3}\!\left(2x^{2}-5\right)^{2}\!\left(5x^{2}-5\right). $$(A1) This can also be written as $20x^{3}(x^{2}-1)(2x^{2}-5)^{2}$.(A1) 也可写作 $20x^{3}(x^{2}-1)(2x^{2}-5)^{2}$。
Product rule with $u=x^{2}+3$, $v=x^{2}-3$: $h'(x)=2x(x^{2}-3)+(x^{2}+3)(2x)=2x^{3}-6x+2x^{3}+6x=4x^{3}$. (M1)取 $u=x^{2}+3$，$v=x^{2}-3$，用乘积法则：$h'(x)=2x(x^{2}-3)+(x^{2}+3)(2x)=2x^{3}-6x+2x^{3}+6x=4x^{3}$。(M1)
Verification by expansion: $h(x)=(x^{2}+3)(x^{2}-3)=x^{4}-9$, so $h'(x)=4x^{3}$. Both give the same answer. (A1)展开验证：$h(x)=(x^{2}+3)(x^{2}-3)=x^{4}-9$，故 $h'(x)=4x^{3}$。两种方法结果一致。(A1)
Differentiate: (a) $f(x)=(3x^{4}-2x+1)^{6}$; (b) $g(x)=\sqrt{5x^{3}+\sqrt{x}}$ (nested chain); (c) $h(x)=\left(\frac{x^{2}+1}{x-1}\right)^{4}$ (chain with inner quotient).
Outer function: $u^{6}$ with $u=3x^{4}-2x+1$. Inner derivative: $u'=12x^{3}-2$. Chain rule: (M1)外层函数为 $u^{6}$，其中 $u=3x^{4}-2x+1$。内层导数：$u'=12x^{3}-2$。链式法则：(M1)
$$ f'(x)=6\!\left(3x^{4}-2x+1\right)^{5}\cdot\!\left(12x^{3}-2\right). $$(A1) This is fully simplified.(A1) 此为最简形式。
Write $g(x)=\left(5x^{3}+x^{1/2}\right)^{1/2}$. The outermost function is $(\cdot)^{1/2}$; its argument is $5x^{3}+x^{1/2}$, which itself contains the inner chain $x^{1/2}$. (M1)将其写为 $g(x)=\left(5x^{3}+x^{1/2}\right)^{1/2}$。最外层函数为 $(\cdot)^{1/2}$；其自变量为 $5x^{3}+x^{1/2}$，其中又含内层链式 $x^{1/2}$。(M1)
Differentiate the outer function and multiply by the derivative of its argument: (M1)对外层函数求导，再乘以自变量的导数：(M1)
$$ g'(x)=\frac{1}{2}\!\left(5x^{3}+x^{1/2}\right)^{-1/2}\cdot\!\left(15x^{2}+\tfrac{1}{2}x^{-1/2}\right). $$This can be written as $\dfrac{15x^{2}+\frac{1}{2}x^{-1/2}}{2\sqrt{5x^{3}+\sqrt{x}}}$. (A1)可写为 $\dfrac{15x^{2}+\frac{1}{2}x^{-1/2}}{2\sqrt{5x^{3}+\sqrt{x}}}$。(A1)
Let $u=\dfrac{x^{2}+1}{x-1}$. Chain rule gives $h'(x)=4u^{3}\cdot u'$. (M1)令 $u=\dfrac{x^{2}+1}{x-1}$。链式法则给出 $h'(x)=4u^{3}\cdot u'$。(M1)
Compute $u'$ by the quotient rule: $u'=\dfrac{2x(x-1)-(x^{2}+1)(1)}{(x-1)^{2}}=\dfrac{2x^{2}-2x-x^{2}-1}{(x-1)^{2}}=\dfrac{x^{2}-2x-1}{(x-1)^{2}}$. (M1)用商法则计算 $u'$：$u'=\dfrac{2x(x-1)-(x^{2}+1)(1)}{(x-1)^{2}}=\dfrac{2x^{2}-2x-x^{2}-1}{(x-1)^{2}}=\dfrac{x^{2}-2x-1}{(x-1)^{2}}$。(M1)
Therefore因此
$$ h'(x)=4\!\left(\frac{x^{2}+1}{x-1}\right)^{3}\cdot\frac{x^{2}-2x-1}{(x-1)^{2}}=\frac{4(x^{2}+1)^{3}(x^{2}-2x-1)}{(x-1)^{5}}. $$(A1)
Differentiate using the quotient rule: (a) $f(x)=\frac{x^{3}-3x}{x^{2}+1}$; (b) $g(x)=\frac{x^{2}-4x+4}{(x+1)^{2}}$; (c) $h(x)=\frac{x^{3}}{x^{2}+3}$, and find all $x$ with $h'(x)=0$.
With $N=x^{3}-3x$ and $D=x^{2}+1$: $N'=3x^{2}-3$, $D'=2x$. Quotient rule $\frac{N'D-ND'}{D^{2}}$: (M1)取 $N=x^{3}-3x$，$D=x^{2}+1$：$N'=3x^{2}-3$，$D'=2x$。商法则 $\frac{N'D-ND'}{D^{2}}$：(M1)
$$ f'(x)=\frac{(3x^{2}-3)(x^{2}+1)-(x^{3}-3x)(2x)}{(x^{2}+1)^{2}}. $$Expand the numerator: $(3x^{4}+3x^{2}-3x^{2}-3)-(2x^{4}-6x^{2})=3x^{4}-3-2x^{4}+6x^{2}=x^{4}+6x^{2}-3$. (A1)展开分子：$(3x^{4}+3x^{2}-3x^{2}-3)-(2x^{4}-6x^{2})=3x^{4}-3-2x^{4}+6x^{2}=x^{4}+6x^{2}-3$。(A1)
$$ f'(x)=\frac{x^{4}+6x^{2}-3}{(x^{2}+1)^{2}}. $$Note $x^{2}-4x+4=(x-2)^{2}$, so $g(x)=\dfrac{(x-2)^{2}}{(x+1)^{2}}$. With $N=(x-2)^{2}$, $D=(x+1)^{2}$: $N'=2(x-2)$, $D'=2(x+1)$. (M1)注意 $x^{2}-4x+4=(x-2)^{2}$，故 $g(x)=\dfrac{(x-2)^{2}}{(x+1)^{2}}$。取 $N=(x-2)^{2}$，$D=(x+1)^{2}$：$N'=2(x-2)$，$D'=2(x+1)$。(M1)
$$ g'(x)=\frac{2(x-2)(x+1)^{2}-(x-2)^{2}\cdot 2(x+1)}{(x+1)^{4}}=\frac{2(x-2)(x+1)\!\left[(x+1)-(x-2)\right]}{(x+1)^{4}}. $$The bracket simplifies to $3$, giving $g'(x)=\dfrac{6(x-2)}{(x+1)^{3}}$. (A1)括号化简为 $3$，得 $g'(x)=\dfrac{6(x-2)}{(x+1)^{3}}$。(A1)
With $N=x^{3}$, $D=x^{2}+3$: $N'=3x^{2}$, $D'=2x$. (M1)取 $N=x^{3}$，$D=x^{2}+3$：$N'=3x^{2}$，$D'=2x$。(M1)
$$ h'(x)=\frac{3x^{2}(x^{2}+3)-x^{3}(2x)}{(x^{2}+3)^{2}}=\frac{3x^{4}+9x^{2}-2x^{4}}{(x^{2}+3)^{2}}=\frac{x^{4}+9x^{2}}{(x^{2}+3)^{2}}=\frac{x^{2}(x^{2}+9)}{(x^{2}+3)^{2}}. $$Since $x^{2}+9\ge 9>0$ and $(x^{2}+3)^{2}>0$ for all real $x$, the only zero of $h'$ is $x^{2}=0$, i.e. $x=0$. (A1)由于对所有实数 $x$ 均有 $x^{2}+9\ge 9>0$ 且 $(x^{2}+3)^{2}>0$，$h'$ 的唯一零点为 $x^{2}=0$，即 $x=0$。(A1)
Prove $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$ using the limit definition, via the add-subtract trick. Then differentiate $p(x)=(x^{3}+1)(2x^{2}-3x)$.
By definition, $(fg)'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$. (M1)由定义，$(fg)'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$。(M1)
Add and subtract $f(x+h)g(x)$ in the numerator (this is the key algebraic trick): (M1)在分子中加减 $f(x+h)g(x)$（这是关键的代数技巧）：(M1)
$$\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}.$$Group and factor: (A1)分组并提取公因子：(A1)
$$=f(x+h)\cdot\frac{g(x+h)-g(x)}{h}+\frac{f(x+h)-f(x)}{h}\cdot g(x).$$(A1) This expresses the single difference quotient as two separate difference quotients, each recognisable as a derivative.(A1) 这将单一差商表示为两个独立差商之和，每个都可识别为某个导数。
Take the limit as $h\to 0$. (M1) The second difference quotient $\dfrac{f(x+h)-f(x)}{h}\to f'(x)$ by definition. The first difference quotient $\dfrac{g(x+h)-g(x)}{h}\to g'(x)$ by definition.令 $h\to 0$ 取极限。(M1) 第二个差商 $\dfrac{f(x+h)-f(x)}{h}\to f'(x)$（由定义）。第一个差商 $\dfrac{g(x+h)-g(x)}{h}\to g'(x)$（由定义）。
The factor $f(x+h)$ in the first term needs care: because $f$ is differentiable at $x$, it is also continuous there, so $f(x+h)\to f(x)$ as $h\to 0$. (R1)第一项中的因子 $f(x+h)$ 需要特别注意：由于 $f$ 在 $x$ 处可微，故在该点也连续，从而当 $h\to 0$ 时 $f(x+h)\to f(x)$。(R1)
$$(fg)'(x)=f(x)\cdot g'(x)+f'(x)\cdot g(x)=f'(x)g(x)+f(x)g'(x). \quad\square$$(A1)
Let $u=x^{3}+1$ and $v=2x^{2}-3x$, so $u'=3x^{2}$ and $v'=4x-3$. (M1)令 $u=x^{3}+1$，$v=2x^{2}-3x$，则 $u'=3x^{2}$，$v'=4x-3$。(M1)
$$p'(x)=3x^{2}\!\left(2x^{2}-3x\right)+\left(x^{3}+1\right)\!\left(4x-3\right).$$Expand: $6x^{4}-9x^{3}$ and $4x^{4}-3x^{3}+4x-3$. (A1)展开：$6x^{4}-9x^{3}$ 与 $4x^{4}-3x^{3}+4x-3$。(A1)
$$p'(x)=6x^{4}-9x^{3}+4x^{4}-3x^{3}+4x-3=10x^{4}-12x^{3}+4x-3.$$(A1)
Prove $\frac{d}{dx}(x^{n})=nx^{n-1}$ for all positive integers $n$ by induction. Then find $f''(x)$ and $f'''(x)$ for $f(x)=x^{5}-3x^{3}+2x$.
Let $f(x)=x^{1}=x$. By the limit definition: (M1)令 $f(x)=x^{1}=x$。由极限定义：(M1)
$$f'(x)=\lim_{h\to 0}\frac{(x+h)-x}{h}=\lim_{h\to 0}\frac{h}{h}=\lim_{h\to 0}1=1.$$Since $1=1\cdot x^{0}$, the formula $nx^{n-1}$ holds for $n=1$. (A1) The base case is established. (R1)由于 $1=1\cdot x^{0}$，公式 $nx^{n-1}$ 对 $n=1$ 成立。(A1) 基础情形已证。(R1)
Assume $\dfrac{d}{dx}(x^{k})=kx^{k-1}$ for some positive integer $k$. (M1) Write $x^{k+1}=x\cdot x^{k}$ and apply the product rule (proved in Q5):假设对某正整数 $k$ 有 $\dfrac{d}{dx}(x^{k})=kx^{k-1}$。(M1) 将 $x^{k+1}=x\cdot x^{k}$，并应用（Q5 中已证的）乘积法则：
$$\frac{d}{dx}\!\left(x^{k+1}\right)=\frac{d}{dx}\!\left(x\cdot x^{k}\right)=1\cdot x^{k}+x\cdot kx^{k-1}=x^{k}+kx^{k}=(k+1)x^{k}.$$(A1) This is exactly the formula $nx^{n-1}$ evaluated at $n=k+1$. (R1)(A1) 这正是公式 $nx^{n-1}$ 在 $n=k+1$ 处的结果。(R1)
By the principle of mathematical induction, $\dfrac{d}{dx}(x^{n})=nx^{n-1}$ holds for all positive integers $n$.由数学归纳法原理，$\dfrac{d}{dx}(x^{n})=nx^{n-1}$ 对所有正整数 $n$ 成立。
$f'(x)=5x^{4}-9x^{2}+2$ (power rule on each term). Differentiate again: (M1)$f'(x)=5x^{4}-9x^{2}+2$（对每项应用幂法则）。再次求导：(M1)
$$f''(x)=20x^{3}-18x.$$Differentiate once more: $f'''(x)=60x^{2}-18$. (A1)再求导一次：$f'''(x)=60x^{2}-18$。(A1)
Derive $Q'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$ from $f=Q\cdot g$ and the product rule. Then differentiate $r(x)=\frac{x^{4}-1}{x^{2}+1}$.
Since $f(x)=Q(x)\cdot g(x)$, differentiate both sides using the product rule: $f'(x)=Q'(x)g(x)+Q(x)g'(x)$. (M1)由 $f(x)=Q(x)\cdot g(x)$，对两边用乘积法则求导：$f'(x)=Q'(x)g(x)+Q(x)g'(x)$。(M1)
Rearrange to isolate $Q'(x)$: $Q'(x)g(x)=f'(x)-Q(x)g'(x)$, so整理以分离 $Q'(x)$：$Q'(x)g(x)=f'(x)-Q(x)g'(x)$，故
$$Q'(x)=\frac{f'(x)-Q(x)g'(x)}{g(x)}=\frac{f'(x)}{g(x)}-\frac{f(x)}{g(x)}\cdot\frac{g'(x)}{g(x)}$$(using $Q=f/g$). (M1) Combine over the common denominator $[g(x)]^{2}$:（利用 $Q=f/g$）。(M1) 通分，公分母为 $[g(x)]^{2}$：
$$Q'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}.\quad\square$$(A1)
Rewrite: $Q'=\dfrac{f'g-fg'}{g^{2}}$. Compare with the product rule $(fg)'=f'g+fg'$. (M1)改写为：$Q'=\dfrac{f'g-fg'}{g^{2}}$。与乘积法则 $(fg)'=f'g+fg'$ 对比。(M1)
The numerator of the quotient rule carries a minus sign before $fg'$, whereas the product rule carries a plus. (A1) The minus arises because differentiating $1/g$ produces $-g'/g^{2}$ (chain rule on the reciprocal), reversing the sign. (R1) A correct mnemonic: "top-prime times bottom, minus top times bottom-prime, over bottom-squared."商法则的分子在 $fg'$ 前有一个负号，而乘积法则有正号。(A1) 负号来源于对 $1/g$ 求导得 $-g'/g^{2}$（对倒数应用链式法则），从而改变了符号。(R1) 正确的记忆口诀：「分子导数乘分母，减去分子乘分母导数，除以分母的平方。」
$N=x^{4}-1$, $D=x^{2}+1$, $N'=4x^{3}$, $D'=2x$. (M1)$N=x^{4}-1$，$D=x^{2}+1$，$N'=4x^{3}$，$D'=2x$。(M1)
$$r'(x)=\frac{4x^{3}(x^{2}+1)-(x^{4}-1)(2x)}{(x^{2}+1)^{2}}=\frac{4x^{5}+4x^{3}-2x^{5}+2x}{(x^{2}+1)^{2}}=\frac{2x^{5}+4x^{3}+2x}{(x^{2}+1)^{2}}.$$Factor: $\dfrac{2x(x^{4}+2x^{2}+1)}{(x^{2}+1)^{2}}=\dfrac{2x(x^{2}+1)^{2}}{(x^{2}+1)^{2}}=2x$. (A1)因式分解：$\dfrac{2x(x^{4}+2x^{2}+1)}{(x^{2}+1)^{2}}=\dfrac{2x(x^{2}+1)^{2}}{(x^{2}+1)^{2}}=2x$。(A1)
This agrees with differentiating $r(x)=(x^{2}-1)(x^{2}+1)/(x^{2}+1)=x^{2}-1$ directly.这与直接对 $r(x)=(x^{2}-1)(x^{2}+1)/(x^{2}+1)=x^{2}-1$ 求导结果一致。
For $F(x)=x^{3}(x^{2}+4)^{5}$: (a) differentiate; (b) factor to the form $x^{2}(x^{2}+4)^{4}(Ax^{2}+B)$; (c) find zeros of $F'$ and classify each.
Let $u=x^{3}$ and $v=(x^{2}+4)^{5}$. Then $u'=3x^{2}$ and, by the chain rule, $v'=5(x^{2}+4)^{4}\cdot 2x=10x(x^{2}+4)^{4}$. (M1)令 $u=x^{3}$，$v=(x^{2}+4)^{5}$。则 $u'=3x^{2}$，由链式法则 $v'=5(x^{2}+4)^{4}\cdot 2x=10x(x^{2}+4)^{4}$。(M1)
$$F'(x)=3x^{2}\!\left(x^{2}+4\right)^{5}+x^{3}\cdot 10x\!\left(x^{2}+4\right)^{4}=3x^{2}\!\left(x^{2}+4\right)^{5}+10x^{4}\!\left(x^{2}+4\right)^{4}.$$(A1)
Both terms share the factor $x^{2}(x^{2}+4)^{4}$. Extract it: (M1)两项共有公因子 $x^{2}(x^{2}+4)^{4}$，提取之：(M1)
$$F'(x)=x^{2}\!\left(x^{2}+4\right)^{4}\!\left[3(x^{2}+4)+10x^{2}\right]=x^{2}\!\left(x^{2}+4\right)^{4}\!\left(3x^{2}+12+10x^{2}\right).$$(A1) Combine like terms in the bracket:(A1) 合并括号内的同类项：
$$F'(x)=x^{2}\!\left(x^{2}+4\right)^{4}\!\left(13x^{2}+12\right).$$So $A=13$ and $B=12$. (A1)故 $A=13$，$B=12$。(A1)
Set $F'(x)=0$: the three factors are $x^{2}$, $(x^{2}+4)^{4}$, and $13x^{2}+12$. (M1)令 $F'(x)=0$：三个因子为 $x^{2}$、$(x^{2}+4)^{4}$ 和 $13x^{2}+12$。(M1)
So the only zero of $F'$ on $\mathbb{R}$ is $x=0$. (A1) For $x\ne 0$: $x^{2}>0$, $(x^{2}+4)^{4}>0$, and $13x^{2}+12>0$, so $F'(x)>0$. Hence $F$ is strictly increasing on $(-\infty,0)$ and on $(0,\infty)$, and $F'$ does not change sign at $x=0$. The point $(0,F(0))=(0,0)$ is neither a local maximum nor a local minimum. (R1)故 $F'$ 在 $\mathbb{R}$ 上的唯一零点为 $x=0$。(A1) 当 $x\ne 0$ 时：$x^{2}>0$，$(x^{2}+4)^{4}>0$，$13x^{2}+12>0$，故 $F'(x)>0$。因此 $F$ 在 $(-\infty,0)$ 和 $(0,\infty)$ 上均严格递增，$F'$ 在 $x=0$ 处不变号。点 $(0,F(0))=(0,0)$ 既非极大值点也非极小值点。(R1)
For $f(x)=\frac{x^{2}-4}{x^{2}+4}$: (a) domain; (b) $f'(x)=\frac{16x}{(x^{2}+4)^{2}}$; (c) horizontal tangent point(s); (d) horizontal asymptotes.
The denominator is $x^{2}+4\ge 4>0$ for all real $x$ (since $x^{2}\ge 0$), so it is never zero. (A1) The domain of $f$ is all of $\mathbb{R}$. (R1)对所有实数 $x$，分母 $x^{2}+4\ge 4>0$（因为 $x^{2}\ge 0$），故分母永不为零。(A1) $f$ 的定义域为全体实数 $\mathbb{R}$。(R1)
$N=x^{2}-4$, $D=x^{2}+4$, $N'=2x$, $D'=2x$. Quotient rule: (M1)$N=x^{2}-4$，$D=x^{2}+4$，$N'=2x$，$D'=2x$。商法则：(M1)
$$f'(x)=\frac{2x(x^{2}+4)-(x^{2}-4)(2x)}{(x^{2}+4)^{2}}.$$Unsimplified numerator: $2x^{3}+8x-2x^{3}+8x=16x$. (A1)未化简的分子：$2x^{3}+8x-2x^{3}+8x=16x$。(A1)
$$f'(x)=\frac{16x}{(x^{2}+4)^{2}}.$$(A1)
$f'(x)=0$ if and only if the numerator $16x=0$, i.e. $x=0$ (the denominator is always positive). (M1)$f'(x)=0$ 当且仅当分子 $16x=0$，即 $x=0$（分母恒为正）。(M1)
At $x=0$: $f(0)=\dfrac{0-4}{0+4}=\dfrac{-4}{4}=-1$. (A1) The only point with a horizontal tangent is $(0,-1)$. (A1)在 $x=0$ 处：$f(0)=\dfrac{0-4}{0+4}=\dfrac{-4}{4}=-1$。(A1) 唯一的水平切线点为 $(0,-1)$。(A1)
As $x\to\pm\infty$, the highest-power terms dominate: (M1)当 $x\to\pm\infty$ 时，最高次项占主导：(M1)
$$f(x)=\frac{x^{2}-4}{x^{2}+4}=\frac{1-4/x^{2}}{1+4/x^{2}}\longrightarrow\frac{1}{1}=1.$$The horizontal asymptote is $y=1$. (A1) To confirm $f$ never equals $1$: set $\dfrac{x^{2}-4}{x^{2}+4}=1$ gives $x^{2}-4=x^{2}+4$, i.e. $-4=4$, a contradiction. So $f(x)<1$ for all $x$, and $f$ only approaches $y=1$.水平渐近线为 $y=1$。(A1) 验证 $f$ 从不等于 $1$：令 $\dfrac{x^{2}-4}{x^{2}+4}=1$ 得 $x^{2}-4=x^{2}+4$，即 $-4=4$，矛盾。故对所有 $x$ 有 $f(x)<1$，$f$ 仅趋近于 $y=1$。
For $G(x)=\frac{(x^{2}+1)^{3}}{x^{3}-8}$ ($x\ne 2$): (a) find $G'(x)$; (b) find $G(3)$, $G'(3)$, and the tangent line at $x=3$; (c) refute the claim that $G'$ can never be negative; (d) identify the two rules misapplied in $G'(x)=\frac{3(x^{2}+1)^{2}}{3x^{2}}$.
Let $N=(x^{2}+1)^{3}$ and $D=x^{3}-8$. Chain rule on $N$: $N'=3(x^{2}+1)^{2}\cdot 2x=6x(x^{2}+1)^{2}$. $D'=3x^{2}$. (M1)令 $N=(x^{2}+1)^{3}$，$D=x^{3}-8$。对 $N$ 应用链式法则：$N'=3(x^{2}+1)^{2}\cdot 2x=6x(x^{2}+1)^{2}$。$D'=3x^{2}$。(M1)
Quotient rule: (M1)商法则：(M1)
$$G'(x)=\frac{6x(x^{2}+1)^{2}(x^{3}-8)-(x^{2}+1)^{3}\cdot 3x^{2}}{(x^{3}-8)^{2}}.$$Factor $(x^{2}+1)^{2}$ from the numerator:从分子中提取 $(x^{2}+1)^{2}$：
$$G'(x)=\frac{(x^{2}+1)^{2}\!\left[6x(x^{3}-8)-3x^{2}(x^{2}+1)\right]}{(x^{3}-8)^{2}}.$$Expand the bracket: $6x^{4}-48x-3x^{4}-3x^{2}=3x^{4}-3x^{2}-48x=3x(x^{3}-x-16)$. (A1)展开括号：$6x^{4}-48x-3x^{4}-3x^{2}=3x^{4}-3x^{2}-48x=3x(x^{3}-x-16)$。(A1)
$$G'(x)=\frac{3x(x^{2}+1)^{2}(x^{3}-x-16)}{(x^{3}-8)^{2}}.$$$G(3)=\dfrac{(9+1)^{3}}{27-8}=\dfrac{1000}{19}$. For $G'(3)$: numerator factors at $x=3$: (M1)$G(3)=\dfrac{(9+1)^{3}}{27-8}=\dfrac{1000}{19}$。计算 $G'(3)$：$x=3$ 处分子的各因子：(M1)
(A1) The tangent line through $\!\left(3,\dfrac{1000}{19}\right)$ with slope $\dfrac{7200}{361}$:(A1) 过点 $\!\left(3,\dfrac{1000}{19}\right)$、斜率为 $\dfrac{7200}{361}$ 的切线：
$$y-\frac{1000}{19}=\frac{7200}{361}(x-3).$$Since $\dfrac{1000}{19}=\dfrac{19000}{361}$, this gives $y=\dfrac{7200}{361}x-\dfrac{21600}{361}+\dfrac{19000}{361}=\dfrac{7200}{361}x-\dfrac{2600}{361}$. (A1)由于 $\dfrac{1000}{19}=\dfrac{19000}{361}$，得 $y=\dfrac{7200}{361}x-\dfrac{21600}{361}+\dfrac{19000}{361}=\dfrac{7200}{361}x-\dfrac{2600}{361}$。(A1)
The sign of $G'$ is determined by the factor $3x(x^{3}-x-16)$, since $(x^{2}+1)^{2}>0$ and $(x^{3}-8)^{2}>0$ always. (M1) At $x=1$: $x^{3}-x-16=1-1-16=-16<0$, and $3x=3>0$, so $3x(x^{3}-x-16)=3\cdot(-16)=-48<0$. Therefore $G'(1)<0$. (A1) The student's error is that a positive numerator of $G$ does not imply a positive derivative of $G$; the denominator and its derivative both contribute.$G'$ 的符号由因子 $3x(x^{3}-x-16)$ 决定，因为 $(x^{2}+1)^{2}>0$ 且 $(x^{3}-8)^{2}>0$ 恒成立。(M1) 在 $x=1$ 处：$x^{3}-x-16=1-1-16=-16<0$，而 $3x=3>0$，故 $3x(x^{3}-x-16)=3\cdot(-16)=-48<0$。因此 $G'(1)<0$。(A1) 该学生的错误在于：$G$ 的分子为正并不意味着 $G$ 的导数为正；分母及其导数均参与影响导数的符号。
Error 1: The student did not apply the quotient rule. They differentiated only the numerator and placed the result over the derivative of the denominator, which is not the quotient rule formula $\frac{N'D-ND'}{D^{2}}$. (A1)错误1：该学生未应用商法则。他们只对分子求导，将结果置于分母导数之上，这不是商法则公式 $\frac{N'D-ND'}{D^{2}}$。(A1)
Error 2: Even if the student had intended some version of "differentiate numerator over derivative of denominator," they placed $3x^{2}$ (the derivative of $x^{3}-8$) in the denominator rather than the denominator itself squared. The correct formula requires $(D)^{2}=(x^{3}-8)^{2}$ in the denominator, not $D'=3x^{2}$. (A1)错误2：即使该学生想用「分子导数除以分母导数」这种形式，他们也将 $3x^{2}$（即 $x^{3}-8$ 的导数）放在了分母，而非分母本身的平方。正确公式要求分母为 $(D)^{2}=(x^{3}-8)^{2}$，而非 $D'=3x^{2}$。(A1)