The derivative of the function f(x) = cos^4 x | vedclass

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(B) Given $f(x) = \cos^4 x + \sin^4 x$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we have $f(x) = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x =

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$\frac{\pi}{4} < x < \frac{\pi}{2}$

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(B) Given $f(x) = \cos^4 x + \sin^4 x$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we have $f(x) = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$. Differentiating with respect to $x$: $f'(x) = 0 - \frac{1}{2} \cdot 2\sin(2x) \cdot \cos(2x) \cdot 2 = -\sin(4x)$. We want $f'(x) > 0$, so $-\sin(4x) > 0$, which implies $\sin(4x) For $0 \le x \le 2\pi$, $0 \le 4x \le 8\pi$. $\sin(4x) For the first interval: $\pi This matches option $B$.

The derivative of the function $f(x) = \cos^4 x + \sin^4 x$ for $0 \le x \le 2\pi$ is positive for:

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