91Ó°ÊÓ

To find the derivative of the function, we need to apply the chain rule which states that \(\frac{d}{dx}(g(f(x))) = g'(f(x)) \cdot f'(x)\). In our case, our inner function \(g(x)\) is the arctangent function, and the outer function \(f(x)\) is the fraction inside the arctangent function. We can write \(f(x)\) as: $$ f(x) = \arctan(u(x)), \textrm{ where } u(x) = \frac{x}{x^2 + 2}. $$ First, let's find the derivative of the inner function \(u(x)\) with respect to \(x\): $$ u'(x) = \frac{(x^2+2)(1)-(x)(2x)}{(x^2+2)^2} = \frac{x^2+2-2x^2}{(x^2+2)^2} = \frac{2-x^2}{(x^2+2)^2}. $$ Now, we know that the derivative of the arctangent function is: $$ \frac{d}{dx}\arctan(u(x)) = \frac{1}{1+u^2(x)}. $$ Using the chain rule, we get: $$ f'(x) = \left(\frac{1}{1+(u(x))^2}\right) \cdot u'(x) = \frac{2-x^2}{(x^2+2)^2} \cdot \frac{1}{1+(\frac{x}{x^2+2})^2}. $$

Simplify the expression of \(f'(x)\) as follows: $$ f'(x) = \frac{2-x^2}{(x^2+2)^2\left(1+(\frac{x}{x^2+2})^2\right)} = \frac{2-x^2}{(x^2+2)^2\left(\frac{(x^2+2)^2+x^2}{(x^2+2)^2}\right)}. $$ Simplifying the denominator, we get: $$ f'(x) = \frac{2-x^2}{(x^2+2)^2+x^2} = \frac{2-x^2}{x^4+4x^2+4+x^2}= \frac{2-x^2}{x^4+5x^2+4}. $$

To find the intervals on which the function is increasing and decreasing, look at the denominator and the numerator of the derivative \(f'(x)\). Since the denominator is always positive, the sign of the derivative will depend on the sign of the numerator. If \(x^2 < 2\), then \(2 - x^2 > 0\), and the function is increasing. If \(x^2 > 2\), then \(2 - x^2 < 0\), and the function is decreasing. In summary, the function is increasing on the interval \((-\sqrt{2},\sqrt{2})\) and decreasing on the intervals \((-\infty,-\sqrt{2})\) and \((\sqrt{2},\infty)\).