Derivatives of Inverse Trig Functions — Question 10

Question 10 - Solution

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

Let $u = \sqrt{1 - x^2}$, then by the chain rule: $$f'(x) = \frac{1}{1 + u^2} \cdot u'$$

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

Step 2: Compute $1 + u^2$: $$1 + u^2 = 1 + (1 - x^2) = 2 - x^2$$

Step 3: Combine everything: $$f'(x) = \frac{1}{1 + u^2} \cdot u' = \frac{1}{2 - x^2} \cdot \frac{-x}{\sqrt{1 - x^2}} = \boxed{ \frac{-x}{(2 - x^2)\sqrt{1 - x^2}} }$$