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If $y=\sin x \cdot \sin 2 x \cdot \sin 3 x \ldots \sin n x$, then $y^{\prime}$ is
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Verified Answer
The correct answer is:
$y \cdot \sum_{k=1}^{n} k \cot k x$
Given, $y=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \ldots . \sin n x$.
Taking log on both sides,
$\log y=\log \sin x+\log \sin 2 x+\ldots+\log \sin n x$
Differentiating w.r.t. $x$,
$$
\frac{1}{y} \cdot \frac{d y}{d x}=1 \cdot \cot x+2 \cot 2 x+\ldots+n \cot n x
$$
$\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=y \cdot \sum_{k=1}^{n} k \cot k x \\ \Rightarrow & y^{\prime}=y \cdot \sum_{k=1}^{n} k \cot k x\end{array}$
Taking log on both sides,
$\log y=\log \sin x+\log \sin 2 x+\ldots+\log \sin n x$
Differentiating w.r.t. $x$,
$$
\frac{1}{y} \cdot \frac{d y}{d x}=1 \cdot \cot x+2 \cot 2 x+\ldots+n \cot n x
$$
$\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=y \cdot \sum_{k=1}^{n} k \cot k x \\ \Rightarrow & y^{\prime}=y \cdot \sum_{k=1}^{n} k \cot k x\end{array}$
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