If f(x) = sin([pi^{2}]x) + cos([-pi^{2}]x),th | vedclass

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(B) We know that $\pi^{2} \approx 9.86$. Since $[\cdot]$ is the greatest integer function: $[\pi^{2}] = [9.86] = 9$. $[-\pi^{2}] = [-9.86] = -10$. Sub

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$9 \cos(9x) - 10 \sin(10x)$

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(B) We know that $\pi^{2} \approx 9.86$. Since $[\cdot]$ is the greatest integer function: $[\pi^{2}] = [9.86] = 9$. $[-\pi^{2}] = [-9.86] = -10$. Substituting these values into the function $f(x)$: $f(x) = \sin(9x) + \cos(-10x)$. Since $\cos(-	heta) = \cos(	heta)$,we have: $f(x) = \sin(9x) + \cos(10x)$. Now,differentiating $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(\sin(9x)) + \frac{d}{dx}(\cos(10x))$. Using the chain rule: $f'(x) = 9 \cos(9x) - 10 \sin(10x)$.

If $f(x) = \sin([\pi^{2}]x) + \cos([-\pi^{2}]x)$,then find $f'(x)$,where $[\cdot]$ denotes the greatest integer function.

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