Derivative of function \(\mathrm{f}(\mathrm{x})\left[\mathrm{x}^{2} /\left(1+\sin ^{2} \mathrm{x}\right)\right]\) is (a) Even function (b) Odd function (c) Not define (d) Increasing Function

To find the derivative of the function, \(f(x) = \frac{x^2}{1+\sin^2 x}\), we will make use of the quotient rule, which states that for any two differentiable functions, u(x) and v(x), the derivative of their quotient is given by: \[\frac{d(u/v)}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\] In our case, \(u(x) = x^2\) and \(v(x) =1 + \sin^2(x)\). We first find the derivatives of both functions: \(u'(x) = \frac{d(x^2)}{dx} = 2x\) \(v'(x) = \frac{d(1+\sin^2 x)}{dx} = 2\sin x\cos x\) Now, we apply the quotient rule:

Using the quotient rule, we have: \[f'(x) = \frac{(2x)(1+\sin^2 x) - (x^2)(2\sin x\cos x)}{(1+\sin^2 x)^2}\] Now, let us simplify the expression and analyze the properties of the obtained derivative.

The given function, \(f'(x)\), can be simplified as: \[f'(x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] Now, let's analyze the properties of this derivative according to the given options: (a) Even function: If \(f'(x) = f'(-x)\), then the function is even. By inserting \(x = -x\), we obtain: \[f'(-x) = \frac{-2x(1+\sin^2 (-x)) + 2x^2\sin (-x)\cos (-x)}{(1+\sin^2 (-x))^2}\] Since \(\sin(-x) = -\sin(x)\) and \(\cos(-x) = \cos(x)\), we can write: \[f'(-x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] However, \(f'(-x) \neq f'(x)\), so the function is not even. (b) Odd function: If \(f'(-x) = -f'(x)\), then try evaluating it for negative x. \[f'(-x) = -\frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] Now, comparing this with \(f'(x)\), we notice that it is not equal to \(-f'(x)\). Therefore, the function is not odd. (c) Not defined: The function is defined for all values of x and has a continuous domain, so it is not undefined. (d) Increasing Function: If the function's derivative is positive for all values of x, then the function is increasing. We have: \[f'(x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] The denominator is always positive. However, we cannot determine if the entire expression is positive or negative. Thus, we cannot conclude that the function is always increasing. Based on our analysis, none of the given options (a), (b), (c), or (d) accurately describe the derivative of the function \(f(x) = \frac{x^2}{1 + \sin^2 x}\). The correct answer would be "None of the above."