Question

y=sinxsin2xsin3x,d2y/dx2=?​

Answer 1

Answer:

$\displaystyle{\dfrac{d^2y}{dx^2}=(-cosec^2 (x) \ -4 \ cosec^2 \ (2x) \ -9 \ cosec^2 \ (3x))}$

Step-by-step explanation:

Given :

y = sin x . sin 2x . sin 3x

Taking log both side

log y = log ( sin x . sin 2x . sin 3x )

Applying log (  mn ) = log m + log n here

log y = log sin x + log sin 2x + log sin 3x

$\displaystyle{We \ know}\\\\\\\displaystyle{(\log x)=\frac{1}{x}}\\\\\\\displaystyle{\dfrac{1}{y}=\log\sin x+\log\sin 2x+\log\sin 3x}$

Now differentiate both side w.r.t.  x

$\displaystyle{\dfrac{1}{y}.\dfrac{dy}{dx} =\log\sin x+\log\sin 2x+\log\sin 3x}\\\\\\\displaystyle{\dfrac{1}{y}.\dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \cos x \ +\dfrac{2}{\sin 2x}\times \cos 2x+ \dfrac{3}{\sin 3x}\times \cos 3x}}\\\\\\\displaystyle{\dfrac{1}{y}.\dfrac{dy}{dx}=\cot x+2\cot2x+3\cot 3x}\\\\\\\displaystyle{\dfrac{dy}{dx}=y\left(\cot x+2\cot2x+3\cot 3x\right)}$

Now ,

$\displaystyle{\dfrac{d^2y}{dx^2}=y\left(\cot x+2\cot2x+3\cot 3x\right)}\\\\\\\displaystyle{\dfrac{d^2y}{dx^2}=\left(\cot x+2\cot2x+3\cot 3x\right)\dfrac{dy}{dx}}\\\\\\\displaystyle{\dfrac{d^2y}{dx^2}=(-cosec^2 (x) \ -4 \ cosec^2 \ (2x) \ -9 \ cosec^2 \ (3x))}$

Thus , we get answer.