The given function is \( \displaystyle y = \frac{\sin^{2}x}{1 + \cos^{2}x} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
We will use the Quotient Rule to solve this derivative. According to the Quotient rule, if \( \displaystyle f(x) = \frac{u(x)}{v(x)} \), then its derivative is: \(\displaystyle f'(x) = \frac{u'(x)v(x) \, – u(x)v'(x)}{[v(x)]^{2}} \).
Thus:
\( \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \left( \frac{\sin^{2}x}{1 + \cos^{2}x} \right) \\ &= \frac{(2\sin x \cdot \cos x)(1 + \cos^{2}x) – (\sin^{2}x)(2\cos x \cdot (-\sin x))}{(1 + \cos^{2}x)^{2}} \\ &= \frac{(\sin 2x)(1 + \cos^{2}x) – (\sin^{2}x)(-\sin 2x)}{(1 + \cos^{2}x)^{2}}, \quad \because \sin 2x = 2\sin x \cdot \cos x \\ &= \frac{\sin 2x (1 + \cos^{2}x + \sin^{2}x)}{(1 + \cos^{2}x)^{2}} \\ &= \frac{2\sin 2x}{(1 + \cos^{2}x)^{2}}, \quad \because \sin^{2}x + \cos^{2}x = 1 \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = \frac{2\sin 2x}{(1 + \cos^{2}x)^{2}}} \\ \)Please let me know in the comments if you find any errors in this solution.