Derivatives of Inverse Trig Functions — Question 6

Question 6 - Solution

We are given: $$f(x) = \tan^{-1}\left( \sqrt{1 - x^2} \right)$$

(a) Finding $f'(x)$:

Let $u(x) = \sqrt{1 - x^2}$, so: $$f(x) = \tan^{-1}(u) \Rightarrow f'(x) = \frac{1}{1 + u^2} \cdot u'(x)$$

First, compute $u'(x)$: $$u(x) = (1 - x^2)^{1/2} \Rightarrow u'(x) = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$$

Now compute: $$f'(x) = \frac{1}{1 + (1 - x^2)} \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{1}{2 - x^2} \cdot \frac{-x}{\sqrt{1 - x^2}}$$

Final Answer: $$\boxed{f'(x) = \frac{-x}{(2 - x^2)\sqrt{1 - x^2}}}$$

(b) Domain of $f'(x)$:

We need: $\sqrt{1 - x^2}$ to be real ⇒ $1 - x^2 \geq 0 \Rightarrow x \in [-1, 1]$ , Denominator $\sqrt{1 - x^2} \neq 0$ ⇒ exclude $x = \pm 1$ , Also ensure $2 - x^2 \neq 0 \Rightarrow x \neq \sqrt{2}, -\sqrt{2}$ (but these are already excluded)

So domain of $f'(x)$ is: $$\boxed{(-1, 1)}$$