Derivatives of Inverse Trig Functions — Question 7

Question 7 - Solution

Domain. Let $s=\sqrt{1+x^2}\ge1$. The argument of $\arcsin$ must satisfy

$$\frac{4x^2}{(s+1)^2}=\frac{4(s-1)}{s+1}\le1.$$

This is equivalent to $s\le5/3$, hence

$$\boxed{D=[-4/3,4/3]}.$$

Test the claimed identity. At $x=4/3$, the arcsine is $\pi/2$, while $\arctan(4/3)<\pi/2$. Thus $f(x)=\arctan x$ is false.

Differentiate. For $|x|<4/3$, put $u=2x/(s+1)$. Then

$$u'=\frac{2}{s(s+1)},\qquad 1-u^2=\frac{5-3s}{s+1}.$$

Consequently

$$\boxed{f'(x)=\frac{2}{s\sqrt{s+1}\sqrt{5-3s}}},\qquad s=\sqrt{1+x^2}.$$

The derivative is defined on $(-4/3,4/3)$ and grows without bound toward either endpoint.