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Question:

If $y=\sum_{k=1}^{6} k \cos ^{-1}\left\{\frac{3}{5} \cos k x-\frac{4}{5} \sin k x\right\}$, then $\frac{d y}{d x}$ at $x=0$ is_____________.

Solution:

$y=\sum_{k=1}^{6} k \cos ^{-1}\left\{\frac{3}{5} \cos k x-\frac{4}{5} \sin k x\right\}$

Let $\cos a=\frac{3}{5}$ and $\sin a=\frac{4}{5}$

$\therefore y=\sum_{k=1}^{6} k \cos ^{-1}\{\cos a \cos k x-\sin a \sin k x\}$

$=\sum_{k=1}^{6} k \cos ^{-1}(\cos (k x+a))=\sum_{k=1}^{6} k(k x+a)=\sum_{k=1}^{6}\left(k^{2} x+a k\right)$

$\therefore \frac{d y}{d x}=\sum_{k=1}^{6} k^{2}=\frac{6(7)(13)}{6}=91$

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