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IAL 2026 Jan FP3 Q9

A Level / Edexcel / FP3

IAL 2026 Jan Paper · Question 9

题目

Problem

y=artanh⁡(x1+x2)y = \operatorname{artanh}\left(\frac{x}{\sqrt{1+x^2}}\right)

(a) Show that

dydx=11+x2\frac{dy}{dx} = \frac{1}{\sqrt{1+x^2}}

(b) Hence, by integrating the result in part (a), prove that

artanh⁡(x1+x2)=arsinh⁡x\operatorname{artanh}\left(\frac{x}{\sqrt{1+x^2}}\right) = \operatorname{arsinh} x
(4)
(2)
题目中文翻译 y=artanh⁡(x1+x2)y = \operatorname{artanh}\left(\frac{x}{\sqrt{1+x^2}}\right)

(a) 证明

dydx=11+x2\frac{dy}{dx} = \frac{1}{\sqrt{1+x^2}}

(b) 由此，积分 (a) 中的结果，证明

artanh⁡(x1+x2)=arsinh⁡x\operatorname{artanh}\left(\frac{x}{\sqrt{1+x^2}}\right) = \operatorname{arsinh} x

解答

(a)

解法一

思路

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令 u=x/1+x2u=x/\sqrt{1+x^2}，先求 uu 的导数，再套用 artanh⁡u\operatorname{artanh}u 的链式求导公式。将 1−u21-u^2 化简后会与 u′u' 中的幂次约去，得到目标式。

答题过程

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Let

u=x1+x2=x(1+x2)−1/2.u=\frac{x}{\sqrt{1+x^2}} =x(1+x^2)^{-1/2}.

Then

dudx= (1+x2)−1/2−x2(1+x2)−3/2= 1(1+x2)3/2.\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} =&\,(1+x^2)^{-1/2} -x^2(1+x^2)^{-3/2}\\ =&\,\frac{1}{(1+x^2)^{3/2}}. \end{align*}

Also,

1−u2= 1−x21+x2= 11+x2.\begin{align*} 1-u^2 =&\,1-\frac{x^2}{1+x^2}\\ =&\,\frac{1}{1+x^2}. \end{align*}

Using the chain rule,

dydx= 11−u2dudx= (1+x2)1(1+x2)3/2= 11+x2,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{1}{1-u^2} \frac{\mathrm{d}u}{\mathrm{d}x}\\ =&\,(1+x^2) \frac{1}{(1+x^2)^{3/2}}\\ =&\,\boxed{\frac{1}{\sqrt{1+x^2}}}, \end{align*}

as required.

解法二

思路

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官方评分资料另给出隐式路线：先对原式两边取 tanh⁡\tanh，再求导。利用 sech⁡2y=1−tanh⁡2y\operatorname{sech}^2y=1-\tanh^2y，可把左边也完全改写成 xx，从而解出导数。

答题过程

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Taking tanh⁡\tanh of both sides gives

tanh⁡y=x1+x2.\tanh y=\frac{x}{\sqrt{1+x^2}}.

Differentiating implicitly,

sech⁡2ydydx=1(1+x2)3/2.\operatorname{sech}^2y \frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{(1+x^2)^{3/2}}.

Since

sech⁡2y= 1−tanh⁡2y= 1−x21+x2= 11+x2,\begin{align*} \operatorname{sech}^2y =&\,1-\tanh^2y\\ =&\,1-\frac{x^2}{1+x^2}\\ =&\,\frac{1}{1+x^2}, \end{align*}

it follows that

11+x2dydx= 1(1+x2)3/2,dydx= 11+x2.\begin{align*} \frac{1}{1+x^2} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{1}{(1+x^2)^{3/2}},\\ \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\boxed{\frac{1}{\sqrt{1+x^2}}}. \end{align*}

(b)

解法一

思路

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“Hence” 要直接积分 (a) 的结果。因为 1/1+x21/\sqrt{1+x^2} 的一个原函数是 arsinh⁡x\operatorname{arsinh}x，两边只可能相差常数；代入 x=0x=0 即可证明该常数为零。

答题过程

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From part (a),

dydx=11+x2.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{\sqrt{1+x^2}}.

Integrating with respect to xx,

y=arsinh⁡x+C.y=\operatorname{arsinh}x+C.

Since

y=artanh⁡(x1+x2),y=\operatorname{artanh} \bigg(\frac{x}{\sqrt{1+x^2}}\bigg),

we have

artanh⁡(x1+x2)=arsinh⁡x+C.\operatorname{artanh} \bigg(\frac{x}{\sqrt{1+x^2}}\bigg) =\operatorname{arsinh}x+C.

For every real xx,

∣x∣<1+x2,|x|<\sqrt{1+x^2},

so the argument of artanh⁡\operatorname{artanh} lies in (−1,1)(-1,1) and both sides are defined. When x=0x=0,

0=0+C,0=0+C,

so C=0C=0. Therefore, for all real xx,

artanh⁡(x1+x2)=arsinh⁡x.\boxed{ \operatorname{artanh} \bigg(\frac{x}{\sqrt{1+x^2}}\bigg) =\operatorname{arsinh}x}.