Question

Differentiate $sin2xcos3x.$

Answer 1

By using the product rule, we have

$\frac{d}{dx}\left(sin2xcos3x\right)=\left(sin2x\right).\frac{d}{dx}.\left(cos3x\right)+\left(cos3x\right).\frac{d}{dx}\left(sin2x\right)$

$=\left(sin2x\right).(-sin3x).3+\left(cos3x\right).\left(cos2x\right).2$ [using chain rule]

=-3~sin~2x~sin~3x~+2~cos~3x~cos~2x