The given function is \( \displaystyle y = \tan\left(\frac{x – x^{-1}}{x + x^{-1}}\right) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Let \( \displaystyle u = \frac{x – x^{-1}}{x + x^{-1}} \). Then: \( y= \tan(u) \).
Simplify the value of u:
\( \begin{aligned} u &= \frac{x – x^{-1}}{x + x^{-1}} = \frac{x – \frac{1}{x}}{x + \frac{1}{x}} = \frac{\frac{x^2 – 1}{x}}{\frac{x^2 + 1}{x}} = \frac{x^2 – 1}{x^2 + 1} \\ \end{aligned} \)Now apply the Chain Rule to find the derivative:
\( \begin{aligned} \frac{dy}{dx} &= \frac{d (\tan(u))}{dx} = \frac{d (\tan(u))}{du} \cdot \frac{du}{dx} = \sec^2(u) \cdot \frac{du}{dx} \\ &= \sec^2(u) \cdot \frac{d}{dx}\left(\frac{x^2 – 1}{x^2 + 1} \right) \\ &= \sec^2(u) \cdot \frac{(2x)(x^2 + 1) – (x^2 – 1)(2x)}{(x^2 + 1)^2} \\ &= \sec^2(u) \cdot \frac{2x^3 + 2x – 2x^3 + 2x}{(x^2 + 1)^2} \\ &= \sec^2(u) \cdot \frac{4x}{(x^2 + 1)^2} \\ &= \sec^2\left(\frac{x^2 – 1}{x^2 + 1}\right) \cdot \frac{4x}{(x^2 + 1)^2} \quad \left( \because u = \frac{x^2 – 1}{x^2 + 1} \right) \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = \frac{4x}{(x^2 + 1)^2} \sec^2\left(\frac{x^2 – 1}{x^2 + 1}\right)} \)Please let me know in the comments if you find any errors in this solution.