the derivative of sin³x.cos3x ?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{following} \mathrm{function}. \\ \mathrm{y}={\sin }^3\mathrm{x}·\cos 3\mathrm{x} \\ \mathrm{Use} \mathrm{product} \mathrm{rule} \mathrm{to} \mathrm{differentiate} \mathrm{the} \mathrm{above} \mathrm{function}. \\ \left(\mathrm{u}·\mathrm{v}\right)\text{'}=\mathrm{u}\text{'}·\mathrm{v}+\mathrm{u}·\mathrm{v}\text{'} \\ \mathrm{This} \mathrm{implies} \mathrm{that}, \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\cos 3\mathrm{x}·\frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^3\mathrm{x}\right)+{\sin }^3\mathrm{x}·\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos 3\mathrm{x}\right) \\ =\cos 3\mathrm{x}·3 {\sin }^2\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}\right)+{\sin }^3\mathrm{x}· \left(-\sin 3\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(3\mathrm{x}\right) \\ =3 \cos 3\mathrm{x} {\sin }^2\mathrm{x} \left(\mathrm{cosx}\right)-{\sin }^3\mathrm{x} \sin 3\mathrm{x} ·\left(3\right) \\ =\mathbf{3}\mathbf{cos}\mathbf{3}\mathbf{x}\mathbf{ }{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{ }\mathbf{cosx}\mathbf{-}\mathbf{3}{\mathbf{sin}}^{\mathbf{3}}\mathbf{x}\mathbf{ }\mathbf{sin}\mathbf{3}\mathbf{x} \end{gathered}$$

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