second order derivative of sin2x.sin3x.sin4x.???

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sin 2\mathrm{x}.\sin 3\mathrm{x}.\sin 4\mathrm{x} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}\left(2\sin 4\mathrm{x}.\sin 3\mathrm{x}\right).\sin 2\mathrm{x} \\ \mathrm{As} 2\mathrm{sinA}.\mathrm{sinB}=\cos \left(\mathrm{A}-\mathrm{B}\right)-\cos \left(\mathrm{A}+\mathrm{B}\right) \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}\left\{\cos \left(4\mathrm{x}-3\mathrm{x}\right)-\cos \left(4\mathrm{x}+3\mathrm{x}\right)\right\}.\sin 2\mathrm{x} \\ =\frac{1}{2}\left\{\mathrm{cosx}-\cos 7\mathrm{x}\right\}.\sin 2\mathrm{x} \\ =\frac{1}{2}\left\{\sin 2\mathrm{x}.\mathrm{cosx}-\sin 2\mathrm{x}.\cos 7\mathrm{x}\right\} \\ =\frac{1}{4}\left\{2\sin 2\mathrm{x}.\mathrm{cosx}-2\sin 2\mathrm{x}.\cos 7\mathrm{x}\right\} \\ \mathrm{As} 2\mathrm{sinA}.\mathrm{cosB}=\sin \left(\mathrm{A}+\mathrm{B}\right)+\sin \left(\mathrm{A}-\mathrm{B}\right) \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{4}\left[\sin \left(2\mathrm{x}+\mathrm{x}\right)+\sin \left(2\mathrm{x}-\mathrm{x}\right)-\left\{\sin \left(2\mathrm{x}+7\mathrm{x}\right)+\sin \left(2\mathrm{x}-7\mathrm{x}\right)\right\}\right] \\ =\frac{1}{4}\left[\sin 3\mathrm{x}+\mathrm{sinx}-\left\{\sin 9\mathrm{x}+\sin \left(-5\mathrm{x}\right)\right\}\right] \\ =\frac{1}{4}\left[\sin 3\mathrm{x}+\mathrm{sinx}-\left\{\sin 9\mathrm{x}-\sin 5\mathrm{x}\right\}\right] \\ =\frac{1}{4}\left[\sin 3\mathrm{x}+\mathrm{sinx}-\sin 9\mathrm{x}+\sin 5\mathrm{x}\right] \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{4}\left[\mathrm{sinx}+\sin 3\mathrm{x}+\sin 5\mathrm{x}-\sin 9\mathrm{x}\right] \\ \mathrm{Differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \mathrm{we} \mathrm{get}: \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{4}\left[\mathrm{cosx}+3\cos 3\mathrm{x}+5\cos 5\mathrm{x}-9\cos 9\mathrm{x}\right] \\ \mathrm{Again} \mathrm{differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \mathrm{we} \mathrm{get}: \\ \Rightarrow \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=\frac{1}{4}\left[\left(-\mathrm{sinx}\right)+3\left(-3\sin 3\mathrm{x}\right)+5\left(-5\sin 5\mathrm{x}\right)-9\left(-9\sin 9\mathrm{x}\right)\right] \\ =\frac{1}{4}\left[81\sin 9\mathrm{x}-25\sin 5\mathrm{x}-9\sin 3\mathrm{x}-\mathrm{sinx}\right] \\ \Rightarrow \boxed{\mathbf{f}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left[\mathbf{81}\mathbf{sin}\mathbf{9}\mathbf{x}\mathbf{-}\mathbf{25}\mathbf{sin}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{9}\mathbf{sin}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{sinx}\right]} \end{gathered}$$

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