Question

d/dx of ( log sin x + log sin 2x + log sin 3x)​

Answer 1

$\bold {\huge {Your ~answer :-}}$

Given Functions —

$y = log( \sin(x) + log( \sin(2x) ) + log( \sin(3x)$

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We know

➡Sin x

dy/dx = cosx

➡log x

dy/dx = 1/x

➡U-V rule(chain rule)

For example

log(2x)

dy/dx = 2/x

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Coming to ur questions-

By u-v Rule and above concept

$\frac{dy}{dx} = \frac{1}{ \sin(x) \ } \times \cos(x) + \frac{2}{ \sin(2x) \ } \times \cos(2x) + \frac{3}{ \sin(3x) \ } \times \cos(3x)$

Since we know cosx/sinx = cotx

dy/dx =cot(x) + 2cot(2x) +3cot(3x)

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Answer 2

Answer:

$\displaystyle{\implies\cot x+2\cot2x+3\cot 3x}$

Step-by-step explanation:

$\displaystyle{\dfrac{d}{dx}=\left(\log\sin x+\log\sin 2x+\log\sin 3x\right)}\\\\\\\displaystyle{We \ know \ \dfrac{d}{dx}(\log x)=\frac{1}{x}}\\\\\\\displaystyle{\dfrac{d}{dx}(\sin x)=\cos x}\\\\\\\displaystyle{Now \ \dfrac{d}{dx}=\left(\log\sin x+\log\sin 2x+\log\sin 3x\right)}\\\\\\\displaystyle{\implies\dfrac{1}{\sin x}\times \cos x \ +\dfrac{2}{\sin 2x}\times \cos 2x+ \dfrac{3}{\sin 3x}\times \cos 3x}$

$\displaystyle{\implies\dfrac{\cos x}{\sin x}+2\times\dfrac{\cos 2x}{\sin 2x}+ 3\times\dfrac{\cos 3x}{\sin 3x}}\\\\\\\\\displaystyle{We \ know \ \frac{\cos x}{\sin x}=\cot x}\\\\\\\displaystyle{\implies\cot x+2\cot2x+3\cot 3x}$

Thus we get answer.