Derivatives of Inverse Trig Functions — Question 2

Question 2 - Solution

(a) Use an identity for tangent of double angle:

We recall: $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$

Let $\theta = \tan^{-1}(x) \Rightarrow \tan(\theta) = x$. Then: $$\tan(2\theta) = \frac{2x}{1 - x^2} \Rightarrow 2\theta = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$$

So: $$f(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) = 2\tan^{-1}(x) \quad \text{for } x \in (-1, 1)$$

(b) Differentiate using identity:

From (a), $f(x) = 2\tan^{-1}(x)$

Use the derivative: $$\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2} \Rightarrow f'(x) = 2 \cdot \frac{1}{1 + x^2} = \boxed{\frac{2}{1 + x^2}}$$