Companion to the University-Style Practice Set配套大学风格练习题集
Sections 1 to 7: $e^x$ and $\ln x$, trig derivatives, general $a^x$ and $\log_a x$, logarithmic differentiation, inverse-trig derivatives, hyperbolic functions, synthesis第 1 至 7 节：$e^x$ 与 $\ln x$、三角导数、一般 $a^x$ 与 $\log_a x$、对数求导法、反三角导数、双曲函数、综合CALC I
Differentiate (a) $f(x)=3e^{x}-5\ln x+2$; (b) $g(x)=4^{x}+\log_{4}x$; (c) $h(x)=x^{2}e^{x}-\ln(x^{3})$.对以下各函数求导 (a) $f(x)=3e^{x}-5\ln x+2$；(b) $g(x)=4^{x}+\log_{4}x$；(c) $h(x)=x^{2}e^{x}-\ln(x^{3})$。
Apply $\frac{d}{dx}e^{x}=e^{x}$ and $\frac{d}{dx}\ln x=\frac{1}{x}$ term by term. (M1) The constant $2$ vanishes.逐项应用 $\frac{d}{dx}e^{x}=e^{x}$ 和 $\frac{d}{dx}\ln x=\frac{1}{x}$。(M1) 常数 $2$ 消去。
$$ f'(x)=3e^{x}-\frac{5}{x}. $$(A1)
The formulas $\frac{d}{dx}a^{x}=a^{x}\ln a$ and $\frac{d}{dx}\log_{a}x=\frac{1}{x\ln a}$ follow from writing both in terms of $e$. (M1)公式 $\frac{d}{dx}a^{x}=a^{x}\ln a$ 和 $\frac{d}{dx}\log_{a}x=\frac{1}{x\ln a}$ 均由将两者用 $e$ 表示后得到。(M1)
$$ g'(x)=4^{x}\ln 4+\frac{1}{x\ln 4}. $$(A1)
Simplify $\ln(x^{3})=3\ln x$ before differentiating. For $x^{2}e^{x}$, apply the product rule: $\frac{d}{dx}[x^{2}e^{x}]=2xe^{x}+x^{2}e^{x}=xe^{x}(x+2)$. (M1)求导前先化简 $\ln(x^{3})=3\ln x$。对 $x^{2}e^{x}$ 应用乘积法则：$\frac{d}{dx}[x^{2}e^{x}]=2xe^{x}+x^{2}e^{x}=xe^{x}(x+2)$。(M1)
$$ h'(x)=xe^{x}(x+2)-\frac{3}{x}. $$(A1)
Differentiate (a) $f(x)=3\sin x-2\cos x+\tan x$; (b) $g(x)=x\cos x+\sec x$; (c) $h(x)=\cot x$ via the quotient rule.对以下各函数求导 (a) $f(x)=3\sin x-2\cos x+\tan x$；(b) $g(x)=x\cos x+\sec x$；(c) 用商法则对 $h(x)=\cot x$ 求导。
Use $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(\tan x)'=\sec^{2}x$ term by term. (M1)逐项应用 $(\sin x)'=\cos x$、$(\cos x)'=-\sin x$、$(\tan x)'=\sec^{2}x$。(M1)
$$ f'(x)=3\cos x+2\sin x+\sec^{2}x. $$(A1)
For $x\cos x$, apply the product rule: $(1)\cos x + x(-\sin x)=\cos x-x\sin x$. Then $(\sec x)'=\sec x\tan x$. (M1)对 $x\cos x$ 应用乘积法则：$(1)\cos x + x(-\sin x)=\cos x-x\sin x$。再有 $(\sec x)'=\sec x\tan x$。(M1)
$$ g'(x)=\cos x-x\sin x+\sec x\tan x. $$(A1)
Write $\cot x=\dfrac{\cos x}{\sin x}$ and apply the quotient rule. (M1)将 $\cot x=\dfrac{\cos x}{\sin x}$ 代入并应用商法则。(M1)
$$ \frac{d}{dx}\cot x=\frac{(-\sin x)\sin x-\cos x(\cos x)}{\sin^{2}x}=\frac{-\sin^{2}x-\cos^{2}x}{\sin^{2}x}=\frac{-1}{\sin^{2}x}=-\csc^{2}x. $$The Pythagorean identity $\sin^{2}x+\cos^{2}x=1$ collapses the numerator. (A1)毕达哥拉斯恒等式 $\sin^{2}x+\cos^{2}x=1$ 化简了分子。(A1)
Differentiate (a) $e^{3x^{2}-1}$; (b) $\ln(\sin x)$; (c) $5^{\cos x}$; (d) $\ln\!\left(\frac{e^{x}+1}{e^{x}-1}\right)$.对以下各函数求导 (a) $e^{3x^{2}-1}$；(b) $\ln(\sin x)$；(c) $5^{\cos x}$；(d) $\ln\!\left(\frac{e^{x}+1}{e^{x}-1}\right)$。
$\frac{d}{dx}e^{3x^{2}-1}=e^{3x^{2}-1}\cdot 6x$. (M1·A1)$\frac{d}{dx}e^{3x^{2}-1}=e^{3x^{2}-1}\cdot 6x$。(M1·A1)
$\frac{d}{dx}\ln(\sin x)=\dfrac{1}{\sin x}\cdot\cos x=\dfrac{\cos x}{\sin x}=\cot x$. (M1·A1)$\frac{d}{dx}\ln(\sin x)=\dfrac{1}{\sin x}\cdot\cos x=\dfrac{\cos x}{\sin x}=\cot x$。(M1·A1)
$\frac{d}{dx}5^{\cos x}=5^{\cos x}\ln 5\cdot(-\sin x)=-5^{\cos x}\sin x\ln 5$. (M1·A1)$\frac{d}{dx}5^{\cos x}=5^{\cos x}\ln 5\cdot(-\sin x)=-5^{\cos x}\sin x\ln 5$。(M1·A1)
Write $q(x)=\ln(e^{x}+1)-\ln(e^{x}-1)$ (log quotient law). (M1) Differentiating each term:将 $q(x)=\ln(e^{x}+1)-\ln(e^{x}-1)$（对数商法则）。(M1) 对各项求导：
$$ q'(x)=\frac{e^{x}}{e^{x}+1}-\frac{e^{x}}{e^{x}-1}=e^{x}\cdot\frac{(e^{x}-1)-(e^{x}+1)}{(e^{x}+1)(e^{x}-1)}=\frac{-2e^{x}}{e^{2x}-1}. $$(A1)
Differentiate (a) $e^{x}\sin x\cos x$; (b) $\dfrac{x^{2}e^{x}}{\ln x}$; (c) $e^{-x^{2}}\tan x$.对以下各函数求导 (a) $e^{x}\sin x\cos x$；(b) $\dfrac{x^{2}e^{x}}{\ln x}$；(c) $e^{-x^{2}}\tan x$。
First use $\sin x\cos x=\tfrac{1}{2}\sin(2x)$ to write $f(x)=\tfrac{1}{2}e^{x}\sin(2x)$. (M1) Apply the product rule: (M1)先用 $\sin x\cos x=\tfrac{1}{2}\sin(2x)$ 将 $f(x)$ 写成 $f(x)=\tfrac{1}{2}e^{x}\sin(2x)$。(M1) 应用乘积法则：(M1)
$$ f'(x)=\tfrac{1}{2}\left[e^{x}\sin(2x)+e^{x}\cdot 2\cos(2x)\right]=\tfrac{1}{2}e^{x}\left[\sin(2x)+2\cos(2x)\right]. $$(A1)
The numerator is $u=x^{2}e^{x}$ and the denominator is $v=\ln x$. Compute $u'=2xe^{x}+x^{2}e^{x}=xe^{x}(2+x)$ and $v'=1/x$. (M1) Quotient rule: (M1)分子为 $u=x^{2}e^{x}$，分母为 $v=\ln x$。计算 $u'=2xe^{x}+x^{2}e^{x}=xe^{x}(2+x)$ 和 $v'=1/x$。(M1) 商法则：(M1)
$$ g'(x)=\frac{xe^{x}(2+x)\cdot\ln x-x^{2}e^{x}\cdot(1/x)}{(\ln x)^{2}}=\frac{xe^{x}(2+x)\ln x-xe^{x}}{(\ln x)^{2}}=\frac{xe^{x}\left[(2+x)\ln x-1\right]}{(\ln x)^{2}}. $$(A1)
Let $u=e^{-x^{2}}$ and $v=\tan x$. Then $u'=e^{-x^{2}}\cdot(-2x)$ and $v'=\sec^{2}x$. (M1)设 $u=e^{-x^{2}}$，$v=\tan x$。则 $u'=e^{-x^{2}}\cdot(-2x)$，$v'=\sec^{2}x$。(M1)
$$ h'(x)=e^{-x^{2}}\sec^{2}x+(-2x)e^{-x^{2}}\tan x=e^{-x^{2}}(\sec^{2}x-2x\tan x). $$(A1)
(a) From $e^{\ln x}=x$, differentiate implicitly to prove $\frac{d}{dx}\ln x=\frac{1}{x}$. (b) From $a^{x}=e^{x\ln a}$, prove $\frac{d}{dx}a^{x}=a^{x}\ln a$.(a) 从 $e^{\ln x}=x$ 出发，通过隐函数求导证明 $\frac{d}{dx}\ln x=\frac{1}{x}$。(b) 从 $a^{x}=e^{x\ln a}$ 出发，证明 $\frac{d}{dx}a^{x}=a^{x}\ln a$。
Start from the identity $e^{\ln x}=x$, valid for all $x>0$. (M1)从恒等式 $e^{\ln x}=x$ 出发，对所有 $x>0$ 成立。(M1)
Differentiate both sides with respect to $x$. The right side gives $1$. The left side requires the chain rule with outer function $e^{u}$ and inner function $u=\ln x$: (M1)对两边关于 $x$ 求导。右边得 $1$。左边需对外层函数 $e^{u}$、内层函数 $u=\ln x$ 应用链式法则：(M1)
$$ \frac{d}{dx}\left[e^{\ln x}\right]=e^{\ln x}\cdot\frac{d}{dx}(\ln x)=x\cdot(\ln x)'. $$Setting this equal to $1$: $x\cdot(\ln x)'=1$, so (A1)令其等于 $1$：$x\cdot(\ln x)'=1$，故 (A1)
$$ \frac{d}{dx}\ln x=\frac{1}{x} \quad \text{for } x>0. $$This derivation is non-circular: we used only $\frac{d}{dx}e^{u}=e^{u}$ (assumed known) and the chain rule. (R1)此推导无循环论证：仅使用了 $\frac{d}{dx}e^{u}=e^{u}$（已知）和链式法则。(R1)
For $a>0$ and $a\ne 1$, write $a^{x}=e^{x\ln a}$ where $\ln a$ is a fixed constant. (M1)对 $a>0$ 且 $a\ne 1$，将 $a^{x}=e^{x\ln a}$ 代入，其中 $\ln a$ 为固定常数。(M1)
Differentiate using the chain rule with outer $e^{u}$ and inner $u=x\ln a$: (M1)用外层 $e^{u}$、内层 $u=x\ln a$ 应用链式法则求导：(M1)
$$ \frac{d}{dx}a^{x}=\frac{d}{dx}e^{x\ln a}=e^{x\ln a}\cdot\frac{d}{dx}(x\ln a)=e^{x\ln a}\cdot\ln a. $$Since $e^{x\ln a}=a^{x}$, this gives (A1)由于 $e^{x\ln a}=a^{x}$，故 (A1)
$$ \frac{d}{dx}a^{x}=a^{x}\ln a. $$When $a=e$ we recover $\frac{d}{dx}e^{x}=e^{x}\cdot 1=e^{x}$, confirming consistency. (R1)当 $a=e$ 时，还原为 $\frac{d}{dx}e^{x}=e^{x}\cdot 1=e^{x}$，验证了一致性。(R1)
(a) From $\sin y=x$ prove $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}$; justify $\cos y>0$. (b) From $\tan y=x$ prove $\frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}$; state the identity used.(a) 从 $\sin y=x$ 证明 $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}$；说明 $\cos y>0$ 的理由。(b) 从 $\tan y=x$ 证明 $\frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}$；说明所用恒等式。
Let $y=\arcsin x$ so that $\sin y=x$ and $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. (M1)设 $y=\arcsin x$，则 $\sin y=x$ 且 $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$。(M1)
Differentiate both sides with respect to $x$ (treating $y$ as a function of $x$): (M1)对两边关于 $x$ 求导（将 $y$ 视为 $x$ 的函数）：(M1)
$$ \cos y\cdot\frac{dy}{dx}=1 \implies \frac{dy}{dx}=\frac{1}{\cos y}. $$We need to express $\cos y$ in terms of $x$. The Pythagorean identity gives $\cos^{2}y=1-\sin^{2}y=1-x^{2}$, so $\cos y=\pm\sqrt{1-x^{2}}$. (A1) On the principal branch $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, cosine is positive, so we take the positive square root. (R1) Therefore需将 $\cos y$ 用 $x$ 表示。毕达哥拉斯恒等式给出 $\cos^{2}y=1-\sin^{2}y=1-x^{2}$，故 $\cos y=\pm\sqrt{1-x^{2}}$。(A1) 在主值域 $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ 上，余弦为正，故取正平方根。(R1) 因此
$$ \frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}, \quad x\in(-1,1). $$(A1)
Let $y=\arctan x$ so that $\tan y=x$ and $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. (M1)设 $y=\arctan x$，则 $\tan y=x$ 且 $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$。(M1)
Differentiate both sides with respect to $x$: (M1)对两边关于 $x$ 求导：(M1)
$$ \sec^{2}y\cdot\frac{dy}{dx}=1 \implies \frac{dy}{dx}=\frac{1}{\sec^{2}y}=\cos^{2}y. $$Apply the Pythagorean identity $\sec^{2}y=1+\tan^{2}y$ (i.e. $1/\cos^{2}y=1+\tan^{2}y$): since $\tan y=x$, (A1)应用毕达哥拉斯恒等式 $\sec^{2}y=1+\tan^{2}y$（即 $1/\cos^{2}y=1+\tan^{2}y$）：由于 $\tan y=x$，(A1)
$$ \sec^{2}y=1+x^{2}. $$This identity holds for all $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, so no sign issue arises. (R1) Therefore此恒等式对所有 $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ 成立，故不存在符号问题。(R1) 因此
$$ \frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}, \quad x\in\mathbb{R}. $$(A1)
(a) From the limit definition and the angle-addition formula, prove $\frac{d}{dx}\sin x=\cos x$; use $\lim_{h\to 0}\frac{\sin h}{h}=1$ and $\lim_{h\to 0}\frac{\cos h-1}{h}=0$. (b) Deduce $\frac{d}{dx}\cos x=-\sin x$ using $\cos x=\sin(\tfrac{\pi}{2}-x)$.(a) 从极限定义和角和公式出发，证明 $\frac{d}{dx}\sin x=\cos x$；使用 $\lim_{h\to 0}\frac{\sin h}{h}=1$ 和 $\lim_{h\to 0}\frac{\cos h-1}{h}=0$。(b) 利用 $\cos x=\sin(\tfrac{\pi}{2}-x)$ 推导 $\frac{d}{dx}\cos x=-\sin x$。
Write the difference quotient directly from the definition: (M1)直接从定义写出差商：(M1)
$$ f'(x)=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}. $$Apply the angle-addition formula $\sin(x+h)=\sin x\cos h+\cos x\sin h$: (M1)应用角和公式 $\sin(x+h)=\sin x\cos h+\cos x\sin h$：(M1)
$$ f'(x)=\lim_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}=\lim_{h\to 0}\left[\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\right]. $$The limit splits across the sum (by the limit sum law). (M1) Applying the given standard limits: (A1)极限可按和式拆分（由极限求和法则）。(M1) 应用已知标准极限：(A1)
$$ f'(x)=\sin x\cdot 0+\cos x\cdot 1=\cos x. $$Hence $\dfrac{d}{dx}\sin x=\cos x$. (A1)故 $\dfrac{d}{dx}\sin x=\cos x$。(A1)
Write $\cos x=\sin\!\left(\tfrac{\pi}{2}-x\right)$ and differentiate using the chain rule with outer function $\sin u$ and inner function $u=\tfrac{\pi}{2}-x$. (M1)将 $\cos x=\sin\!\left(\tfrac{\pi}{2}-x\right)$ 代入，用外层函数 $\sin u$、内层函数 $u=\tfrac{\pi}{2}-x$ 应用链式法则求导。(M1)
$$ \frac{d}{dx}\cos x=\frac{d}{dx}\sin\!\left(\tfrac{\pi}{2}-x\right)=\cos\!\left(\tfrac{\pi}{2}-x\right)\cdot(-1). $$(A1) By the cofunction identity $\cos\!\left(\tfrac{\pi}{2}-x\right)=\sin x$, so $\dfrac{d}{dx}\cos x=-\sin x$. (R1)(A1) 由余函数恒等式 $\cos\!\left(\tfrac{\pi}{2}-x\right)=\sin x$，故 $\dfrac{d}{dx}\cos x=-\sin x$。(R1)
(a) Find $\frac{dy}{dx}$ for $y=x^{x}$ ($x>0$). (b) Find $\frac{dy}{dx}$ for $y=\frac{x^{3}\sqrt{x+1}}{e^{2x}(1+x^{2})}$ using logarithmic differentiation.(a) 求 $y=x^{x}$（$x>0$）的 $\frac{dy}{dx}$。(b) 用对数求导法求 $y=\frac{x^{3}\sqrt{x+1}}{e^{2x}(1+x^{2})}$ 的 $\frac{dy}{dx}$。
Since both the base and the exponent contain $x$, neither the power rule nor the exponential rule applies directly. (M1) Take the natural logarithm of both sides:由于底数和指数均含 $x$，幂法则和指数法则均不能直接应用。(M1) 对两边取自然对数：
$$ \ln y=\ln\!\left(x^{x}\right)=x\ln x. $$Differentiate both sides with respect to $x$ (using $\frac{d}{dx}\ln y=\frac{1}{y}\frac{dy}{dx}$ on the left and the product rule on the right): (M1)对两边关于 $x$ 求导（左边用 $\frac{d}{dx}\ln y=\frac{1}{y}\frac{dy}{dx}$，右边用乘积法则）：(M1)
$$ \frac{1}{y}\frac{dy}{dx}=\ln x+x\cdot\frac{1}{x}=\ln x+1. $$Multiply both sides by $y=x^{x}$: (A1)两边乘以 $y=x^{x}$：(A1)
$$ \frac{dy}{dx}=x^{x}(\ln x+1). $$(A1) At $x=1$ this gives $1^{1}(0+1)=1$, a useful sanity check.(A1) 当 $x=1$ 时得 $1^{1}(0+1)=1$，可作为验证。
Take the natural logarithm and apply log laws to convert products/quotients/powers into sums/differences: (M1)取自然对数并应用对数法则，将乘积、商式、幂次转化为和差：(M1)
$$ \ln y=3\ln x+\tfrac{1}{2}\ln(x+1)-2x-\ln(1+x^{2}). $$Differentiate both sides with respect to $x$: (M1)对两边关于 $x$ 求导：(M1)
$$ \frac{1}{y}\frac{dy}{dx}=\frac{3}{x}+\frac{1}{2(x+1)}-2-\frac{2x}{1+x^{2}}. $$(A1) Multiply through by $y$: (A1)(A1) 两边乘以 $y$：(A1)
$$ \frac{dy}{dx}=\frac{x^{3}\sqrt{x+1}}{e^{2x}(1+x^{2})}\left[\frac{3}{x}+\frac{1}{2(x+1)}-2-\frac{2x}{1+x^{2}}\right]. $$(a) Prove $\frac{d}{dx}\sinh x=\cosh x$ and $\frac{d}{dx}\cosh x=\sinh x$ from the definitions. (b) Differentiate $f(x)=\sinh(x^{2})\cdot e^{x}$. (c) Find and classify all critical points of $g(x)=xe^{-x}$.(a) 从定义证明 $\frac{d}{dx}\sinh x=\cosh x$ 和 $\frac{d}{dx}\cosh x=\sinh x$。(b) 对 $f(x)=\sinh(x^{2})\cdot e^{x}$ 求导。(c) 求并分类 $g(x)=xe^{-x}$ 的所有极值点。
Recall $\sinh x=\dfrac{e^{x}-e^{-x}}{2}$ and $\cosh x=\dfrac{e^{x}+e^{-x}}{2}$. Differentiate each using $\frac{d}{dx}e^{-x}=-e^{-x}$: (M1)回忆 $\sinh x=\dfrac{e^{x}-e^{-x}}{2}$ 和 $\cosh x=\dfrac{e^{x}+e^{-x}}{2}$。利用 $\frac{d}{dx}e^{-x}=-e^{-x}$ 对各式求导：(M1)
$$ \frac{d}{dx}\sinh x=\frac{e^{x}-(-e^{-x})}{2}=\frac{e^{x}+e^{-x}}{2}=\cosh x. \quad\text{(A1)} $$ $$ \frac{d}{dx}\cosh x=\frac{e^{x}+(-e^{-x})}{2}=\frac{e^{x}-e^{-x}}{2}=\sinh x. \quad\text{(A1)} $$Let $u=\sinh(x^{2})$ and $v=e^{x}$. Then $u'=\cosh(x^{2})\cdot 2x$ (chain rule) and $v'=e^{x}$. (M1) Product rule: (M1)设 $u=\sinh(x^{2})$，$v=e^{x}$。则 $u'=\cosh(x^{2})\cdot 2x$（链式法则），$v'=e^{x}$。(M1) 乘积法则：(M1)
$$ f'(x)=\cosh(x^{2})\cdot 2x\cdot e^{x}+\sinh(x^{2})\cdot e^{x}=e^{x}\!\left[2x\cosh(x^{2})+\sinh(x^{2})\right]. $$(A1)
Differentiate using the product rule: $g'(x)=e^{-x}+x\cdot(-e^{-x})=e^{-x}(1-x)$. (M1) Since $e^{-x}>0$ for all $x$, we have $g'(x)=0$ if and only if $1-x=0$, i.e. $x=1$. (A1)用乘积法则求导：$g'(x)=e^{-x}+x\cdot(-e^{-x})=e^{-x}(1-x)$。(M1) 由于对所有 $x$ 均有 $e^{-x}>0$，故 $g'(x)=0$ 当且仅当 $1-x=0$，即 $x=1$。(A1)
Compute the second derivative: $g''(x)=\frac{d}{dx}[e^{-x}(1-x)]=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}(x-2)$. (M1) At $x=1$: $g''(1)=e^{-1}(1-2)=-e^{-1}<0$. By the second-derivative test, $x=1$ is a local maximum. (A1)计算二阶导数：$g''(x)=\frac{d}{dx}[e^{-x}(1-x)]=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}(x-2)$。(M1) 在 $x=1$ 处：$g''(1)=e^{-1}(1-2)=-e^{-1}<0$。由二阶导数判别法，$x=1$ 是极大值点。(A1)
The maximum value is $g(1)=1\cdot e^{-1}=e^{-1}$.极大值为 $g(1)=1\cdot e^{-1}=e^{-1}$。
(a) Differentiate $f(x)=\arctan\!\left(\frac{2x}{1-x^{2}}\right)$ and simplify. (b) Differentiate $p(x)=x\arcsin x+\sqrt{1-x^{2}}$ and simplify fully. (c) Find the tangent line to $y=e^{\arctan x}$ at $x=1$.(a) 对 $f(x)=\arctan\!\left(\frac{2x}{1-x^{2}}\right)$ 求导并化简。(b) 对 $p(x)=x\arcsin x+\sqrt{1-x^{2}}$ 求导并完全化简。(c) 求 $y=e^{\arctan x}$ 在 $x=1$ 处的切线。
Let $u=\dfrac{2x}{1-x^{2}}$. The chain rule gives $f'(x)=\dfrac{1}{1+u^{2}}\cdot u'$. (M1)设 $u=\dfrac{2x}{1-x^{2}}$。链式法则给出 $f'(x)=\dfrac{1}{1+u^{2}}\cdot u'$。(M1)
Compute $u'$ by the quotient rule: $u'=\dfrac{2(1-x^{2})-2x(-2x)}{(1-x^{2})^{2}}=\dfrac{2+2x^{2}}{(1-x^{2})^{2}}=\dfrac{2(1+x^{2})}{(1-x^{2})^{2}}$. (M1)用商法则计算 $u'$：$u'=\dfrac{2(1-x^{2})-2x(-2x)}{(1-x^{2})^{2}}=\dfrac{2+2x^{2}}{(1-x^{2})^{2}}=\dfrac{2(1+x^{2})}{(1-x^{2})^{2}}$。(M1)
Compute $1+u^{2}$: $u^{2}=\dfrac{4x^{2}}{(1-x^{2})^{2}}$, so计算 $1+u^{2}$：$u^{2}=\dfrac{4x^{2}}{(1-x^{2})^{2}}$，故
$$ 1+u^{2}=\frac{(1-x^{2})^{2}+4x^{2}}{(1-x^{2})^{2}}=\frac{1-2x^{2}+x^{4}+4x^{2}}{(1-x^{2})^{2}}=\frac{(1+x^{2})^{2}}{(1-x^{2})^{2}}. \quad\text{(A1)} $$Therefore因此
$$ f'(x)=\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}\cdot\frac{2(1+x^{2})}{(1-x^{2})^{2}}=\frac{2}{1+x^{2}}. \quad\text{(A1)} $$Apply the product rule to $x\arcsin x$: $\arcsin x + \dfrac{x}{\sqrt{1-x^{2}}}$. (M1) Differentiate $\sqrt{1-x^{2}}=(1-x^{2})^{1/2}$ by the chain rule: $\dfrac{-2x}{2\sqrt{1-x^{2}}}=\dfrac{-x}{\sqrt{1-x^{2}}}$. (M1) Add the two pieces:对 $x\arcsin x$ 应用乘积法则：$\arcsin x + \dfrac{x}{\sqrt{1-x^{2}}}$。(M1) 用链式法则对 $\sqrt{1-x^{2}}=(1-x^{2})^{1/2}$ 求导：$\dfrac{-2x}{2\sqrt{1-x^{2}}}=\dfrac{-x}{\sqrt{1-x^{2}}}$。(M1) 将两部分相加：
$$ p'(x)=\arcsin x+\frac{x}{\sqrt{1-x^{2}}}-\frac{x}{\sqrt{1-x^{2}}}=\arcsin x. \quad\text{(A1)} $$The two rational terms cancel exactly.两个有理项恰好消去。
Let $r(x)=e^{\arctan x}$. At $x=1$: $\arctan 1=\dfrac{\pi}{4}$, so $r(1)=e^{\pi/4}$. The point of tangency is $\left(1,e^{\pi/4}\right)$. (M1)设 $r(x)=e^{\arctan x}$。在 $x=1$ 处：$\arctan 1=\dfrac{\pi}{4}$，故 $r(1)=e^{\pi/4}$。切点为 $\left(1,e^{\pi/4}\right)$。(M1)
Differentiate by the chain rule: $r'(x)=e^{\arctan x}\cdot\dfrac{1}{1+x^{2}}$. At $x=1$: $r'(1)=\dfrac{e^{\pi/4}}{2}$. (A1)用链式法则求导：$r'(x)=e^{\arctan x}\cdot\dfrac{1}{1+x^{2}}$。在 $x=1$ 处：$r'(1)=\dfrac{e^{\pi/4}}{2}$。(A1)
Point-slope form: $y-e^{\pi/4}=\dfrac{e^{\pi/4}}{2}(x-1)$, which simplifies to (M1)点斜式：$y-e^{\pi/4}=\dfrac{e^{\pi/4}}{2}(x-1)$，化简得 (M1)
$$ y=\frac{e^{\pi/4}}{2}(x+1). $$