Let $f(x) = \frac{\cos^2 x}{1 + \sin^2 x}$. Evaluate the value of $3f(\pi/4) - 9f'(\pi/4)$.

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Professor PC Thomas Classes · Answer

First, we evaluate $f(\pi/4)$: $f(\pi/4) = \frac{\cos^2(\pi/4)}{1 + \sin^2(\pi/4)} = \frac{1/2}{1 + 1/2} = \frac{1/2}{3/2} = \frac{1}{3}$. Next, we simplify $f(x)$ to find its derivative more easily: $f(x) = \frac{1 - \sin^2 x}{1 + \sin^2 x} = \frac{2 - (1 + \sin^2 x)}{1 + \sin^2 x} = \frac{2}{1 + \sin^2 x} - 1$. Differentiating with respect to $x$, we get: $f'(x) = -\frac{2}{(1 + \sin^2 x)^2} \cdot (2\sin x \cos x) = -\frac{2\sin 2x}{(1 + \sin^2 x)^2}$. Now, evaluate $f'(\pi/4)$: $f'(\pi/4) = -\frac{2\sin(\pi/2)}{(1 + \sin^2(\pi/4))^2} = -\frac{2(1)}{(1 + 1/2)^2} = -\frac{2}{9/4} = -\frac{8}{9}$. Finally, substitute these values into the given expression: $3f(\pi/4) - 9f'(\pi/4) = 3\left(\frac{1}{3}\right) - 9\left(-\frac{8}{9}\right) = 1 + 8 = 9$.