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 \(\displaystyle u = \frac{1 – x^2}{1 + x^2} \), then \( y = \cos(u) \).
First, find \(\displaystyle \frac{du}{dx} \) using the quotient rule:
\( \displaystyle \frac{du}{dx} = \frac{(1 + x^2)(-2x)\, – (1\, – x^2)(2x)}{(1 + x^2)^2} \\ \displaystyle = \frac{-2x(1 + x^2 + 1\, – x^2)}{(1 + x^2)^2} \\ \displaystyle = \frac{-4x}{(1 + x^2)^2} \\ \)Now, apply the chain rule to find \(\displaystyle \frac{dy}{dx} \):
\( \displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \\ \displaystyle = -\sin(u) \cdot \frac{-4x}{(1 + x^2)^2} \\ \displaystyle = \sin\left(\frac{1\, – x^2}{1 + x^2}\right) \cdot \frac{4x}{(1 + x^2)^2} \\ \)Thus, the derivative of the function is:
\( \displaystyle \boxed{\frac{dy}{dx} = \frac{4x}{(1 + x^2)^2} \cdot \sin\left(\frac{1\, – x^2}{1 + x^2}\right)} \\ \)Please let me know in the comments if you find any errors in this solution.