y=e^(ax)·cos³x·sin²x Use the product rule for calculating the derivatives of product-form-functions: if y=u·v·w and u,v,w are functions of x, y'=(u)'·v·w+u·(v)'·w+u·v·(w)'
Therfore, y'=(e^(ax))'·cos³x·sin²x+e^(ax)·(cos³x)'·sin²x+e^ax·cos³x·(sin²x)' ··· Eq.1
Here, use the chain rule for the differentiation of composite functions: if y=f(u) and u=f(x), dy/dx=dy/du·du/dx
Therfore Eq.1 can be restated: y=(e^(ax)·1a)·cos³x·sin²x+e^(ax)·(3cos²x·(-sinx))·sin²x+e^(ax)·cos³x·(2sinx·cosx)=e^(ax)·cos²x·sinx·(a·cosx·sinx-3sin²x+2cos²x)
Answer: dy/dx=e^(ax)·cos²xsinx·(a·cosx·sinx+2cos²x-3sin²x)