The given function is \( \displaystyle y = \cos\left(\frac{1\,- x^2}{1 + x^2}\right) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Let \( \displaystyle u = \frac{1\,- x^2}{1 + x^2} \), so \( y = \cos(u) \). Then:
\( \begin{aligned} &\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \\ &\text{First compute: } \frac{dy}{du} \\ &\frac{dy}{du} = \frac{d}{du}\cos(u) = -\sin(u) = -\sin\left(\frac{1 – x^2}{1 + x^2}\right) \\ &\text{Now compute: } \frac{du}{dx} \\ &\frac{du}{dx} = \frac{d}{dx}\left(\frac{1-x^2}{1 + x^2} \right) \\ &= \frac{(-2x)(1 + x^2)\,- (1 – x^2)(2x)}{(1 + x^2)^2} \\ &= \frac{-2x\,- 2x^3\, -2x + 2x^3}{(1 + x^2)^2} \\ &= \frac{-4x}{(1 + x^2)^2} \\ &\text{By combining results, we get:} \\ &\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin\left(\frac{1 – x^2}{1 + x^2}\right) \cdot \left(\frac{-4x}{(1 + x^2)^2}\right) \\ &= \frac{4x}{(1 + x^2)^2} \cdot \sin\left(\frac{1 – x^2}{1 + x^2}\right) \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = \frac{4x}{(1 + x^2)^2} \cdot \sin\left(\frac{1 – x^2}{1 + x^2}\right)} \\ \)Please let me know in the comments if you find any errors in this solution.