Question

If $ y=\sin x\sin 2x\sin 3x........\sin nx $ , then $ y' $ is
A. $ \sum\limits_{k=1}^{n}{k\tan kx} $
B. $ y\sum\limits_{k=1}^{n}{k\cot kx} $
C. $ y\sum\limits_{k=1}^{n}{k\tan kx} $
D. $ \sum\limits_{k=1}^{n}{\cot kx} $

Answer 1

Hint: We solve the given equation using the identity formula of logarithm where the base of $ \ln $ is always $ e $ . The first step would be to take a summation form. Then we first define the derivative rule and how the differentiation of function works.

Complete step-by-step answer:
We have $ y =\sin x\sin 2x\sin 3x........\sin nx $ .
We take logarithm both sides to use the form $ \log \left( ab \right)=\log a+\log b $
 $ \begin{align}
  & \log y \\
 & =\log \left( \sin x\sin 2x\sin 3x........\sin nx \right) \\
 & =\log \left( \sin x \right)+\log \left( \sin 2x \right)+\log \left( \sin 3x \right).....+\log \left( \sin nx \right) \\
\end{align} $
We differentiate the given function $ \log y=\log \left( \sin x \right)+\log \left( \sin 2x \right)+.....+\log \left( \sin nx \right) $ with respect to $ x $ using the chain rule.
We now discuss the multiplication process of two functions where \[f\left( x \right)=u\left( x \right)v\left( x \right)\]
Differentiating \[f\left( x \right)=uv\], we get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].
\[\begin{align}
  & \log y=\log \left( \sin x \right)+\log \left( \sin 2x \right)+.....+\log \left( \sin nx \right) \\
 & \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}+2\dfrac{\cos 2x}{\sin 2x}+....+n\dfrac{\cos nx}{\sin nx} \\
 & \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\cot x+2\cot 2x+....+n\cot nx=\sum\limits_{k=1}^{n}{k\cot kx} \\
 & \Rightarrow \dfrac{dy}{dx}=y\sum\limits_{k=1}^{n}{k\cot kx} \\
\end{align}\]
Therefore, the differentiation of $ y=\sin x\sin 2x\sin 3x........\sin nx $ is $ y\sum\limits_{k=1}^{n}{k\cot kx} $ . The correct option is B.
So, the correct answer is “Option B”.

Note: We need remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t cancelling out the part \[d\left[ h\left( x \right) \right]\]. Canceling the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate.