Derivatives of Inverse Trig Functions — Question 9

Question 9 - Solution

We are given: $$f(x) = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$$

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

Step 1: Differentiate $u = \frac{2x}{1 + x^2}$:

Using the quotient rule: $$u' = \frac{(1 + x^2)(2) - 2x(2x)}{(1 + x^2)^2} = \frac{2(1 + x^2) - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$$

Step 2: Simplify $\sqrt{1 - u^2}$:

$$u = \frac{2x}{1 + x^2} \Rightarrow u^2 = \frac{4x^2}{(1 + x^2)^2} \Rightarrow 1 - u^2 = 1 - \frac{4x^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2}$$

$$= \frac{1 + 2x^2 + x^4 - 4x^2}{(1 + x^2)^2} = \frac{1 - 2x^2 + x^4}{(1 + x^2)^2} = \frac{(1 - x^2)^2}{(1 + x^2)^2} \Rightarrow \sqrt{1 - u^2} = \frac{|1 - x^2|}{1 + x^2}$$

Since $-1 < x < 1$, we have $1 - x^2 > 0$, so: $$\sqrt{1 - u^2} = \frac{1 - x^2}{1 + x^2}$$

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

Cancel $1 - x^2$ and one factor of $1 + x^2$: $$f'(x) = \frac{2}{1 + x^2}$$

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