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Question: If \( y=\sin x\sin 2x\sin 3x........\sin nx \) , then \( y' \) is A. \( \sum\limits_{k=1}^{n}{k\ta...

If y=sin⁡xsin⁡2xsin⁡3x........sin⁡nxy=\sin x\sin 2x\sin 3x........\sin nx , then y′y' is
A. ∑k=1nktan⁡kx\sum\limits_{k=1}^{n}{k\tan kx}
B. y∑k=1nkcot⁡kxy\sum\limits_{k=1}^{n}{k\cot kx}
C. y∑k=1nktan⁡kxy\sum\limits_{k=1}^{n}{k\tan kx}
D. ∑k=1ncot⁡kx\sum\limits_{k=1}^{n}{\cot kx}

Explanation

Solution

Hint : We solve the given equation using the identity formula of logarithm where the base of ln⁡\ln is always ee . 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⁡xsin⁡2xsin⁡3x........sin⁡nxy =\sin x\sin 2x\sin 3x........\sin nx .
We take logarithm both sides to use the form log⁡(ab)=log⁡a+log⁡b\log \left( ab \right)=\log a+\log b
log⁡y =log⁡(sin⁡xsin⁡2xsin⁡3x........sin⁡nx) =log⁡(sin⁡x)+log⁡(sin⁡2x)+log⁡(sin⁡3x).....+log⁡(sin⁡nx) \begin{aligned} & \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{aligned}
We differentiate the given function log⁡y=log⁡(sin⁡x)+log⁡(sin⁡2x)+.....+log⁡(sin⁡nx)\log y=\log \left( \sin x \right)+\log \left( \sin 2x \right)+.....+\log \left( \sin nx \right) with respect to xx using the chain rule.
We now discuss the multiplication process of two functions where f(x)=u(x)v(x)f\left( x \right)=u\left( x \right)v\left( x \right)
Differentiating f(x)=uvf\left( x \right)=uv, we get ddx[f(x)]=ddx[uv]=udvdx+vdudx\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}.

& \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{aligned}$$ Therefore, the differentiation of $ y=\sin x\sin 2x\sin 3x........\sin nx $ is $ y\sum\limits_{k=1}^{n}{k\cot kx} $ . The correct option is B. **So, the correct answer is “Option B”.** **Note** : We need remember that in the chain rule $$\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.