An inverse-trigonometric identity and its endpoints · source reference

Printed question

(i) Show that $\frac{d}{dx}\left(\cos^{-1}\frac{2x}{1+x^2}\right)=\frac{-2}{1+x^2}$ for $-1\le x\le1$. Explain, in your working, how you use the condition $-1\le x\le1$ to conclude your answer. [3]

(ii) Hence, determine the constants $A$ and $B$ satisfying the equation $\cos^{-1}\frac{2x}{1+x^2}+A\tan^{-1}x=B$, for $-1\le x\le1$. [3]

[2009/YJC/Prelim/I/5(a)]

Printed working

(i)

$$\begin{aligned} \frac{d}{dx}\left(\cos^{-1}\frac{2x}{1+x^2}\right) &=\frac{-1}{\sqrt{1-(\frac{2x}{1+x^2})^2}}\left(\frac{(1+x^2)(2)-(2x)(2x)}{(1+x^2)^2}\right)\\ &=\frac{-(1+x^2)}{\sqrt{(1+x^2)^2-4x^2}}\left(\frac{2+2x^2-4x^2}{(1+x^2)^2}\right)\\ &=\frac{-(2-2x^2)}{\sqrt{1-2x^2+x^4}}\left(\frac1{1+x^2}\right)\\ &=\frac{-2(1-x^2)}{\sqrt{(1-x^2)^2}}\left(\frac1{1+x^2}\right). \end{aligned}$$

Note that

$$\sqrt{(1-x^2)^2}=\begin{cases}1-x^2,&1-x^2\ge0,\\-(1-x^2),&1-x^2<0\end{cases}=\begin{cases}1-x^2,&-1\le x\le1,\\-(1-x^2),&x<-1\text{ or }x>1.\end{cases}$$

Since it is given that $-1\le x\le1$, the printed cancellation gives $\frac{-2(1-x^2)}{1-x^2}\frac1{1+x^2}=\frac{-2}{1+x^2}$ (shown).

(ii) Differentiating the given identity with respect to $x$,

$$\frac{-2}{1+x^2}+\frac{A}{1+x^2}=0\Rightarrow\frac{A-2}{1+x^2}=0\Rightarrow A-2=0\Rightarrow A=2.$$

Thus $\cos^{-1}\frac{2x}{1+x^2}+2\tan^{-1}x=B$. Substituting $x=0$ (which satisfies $-1\le x\le1$) gives $\cos^{-1}(0)+2\tan^{-1}0=B\Rightarrow B=\frac\pi2$.

Endpoint qualification

The ordinary two-sided derivative computation applies on $-1<x<1$. Cancellation at $x=\pm1$ divides by zero; the restricted identity is continuous on the closed interval and has the stated one-sided derivatives at its endpoints. The global composite has a cusp there, so a two-sided derivative at either endpoint does not exist.

Teacher notes

From MF26: $\frac{d}{dx}(\cos^{-1}x)=-\frac1{\sqrt{1-x^2}}$.

Apply Chain Rule: let $u=\frac{2x}{1+x^2}$. Then $\frac{d}{dx}(\cos^{-1}u)=\frac{d}{du}(\cos^{-1}u)\frac{du}{dx}=-\frac1{\sqrt{1-u^2}}\frac{du}{dx}=-\frac1{\sqrt{1-(\frac{2x}{1+x^2})^2}}\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)$. Apply quotient rule to this last derivative.

Answer key

(ii) $A=2$, $B=\frac\pi2$.