If f(x) = (cos x)(cos 2x) ldots (cos nx), the | vedclass

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(B) Given $f(x) = \prod_{r=1}^n \cos(rx)$. Taking the natural logarithm on both sides, we get $\ln|f(x)| = \sum_{r=1}^n \ln|\cos(rx)|$. Differentiatin

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(B) Given $f(x) = \prod_{r=1}^n \cos(rx)$. Taking the natural logarithm on both sides, we get $\ln|f(x)| = \sum_{r=1}^n \ln|\cos(rx)|$. Differentiating both sides with respect to $x$, we have $\frac{f^{\prime}(x)}{f(x)} = \sum_{r=1}^n \frac{1}{\cos(rx)} \cdot (-\sin(rx) \cdot r) = -\sum_{r=1}^n r 	an(rx)$. Multiplying both sides by $f(x)$, we get $f^{\prime}(x) = -f(x) \sum_{r=1}^n r 	an(rx)$. Rearranging the terms, we obtain $f^{\prime}(x) + \sum_{r=1}^n (r 	an rx) f(x) = 0$.

If $f(x) = (\cos x)(\cos 2x) \ldots (\cos nx)$, then $f^{\prime}(x) + \sum_{r=1}^n (r \tan rx) f(x)$ is equal to

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