Question: If y=e^(ax)cos^3xsin^2x find dy/dx.
The expression is,
y=e^(ax)cos^3(x).sin^2(x)
Use the product rule, twice.
Let y = uv
then y' = u'v + uv'
where u = e^(ax) and u' = a.e^(ax)
v = cos^3(x).sin^2(x) and v' = (cos^3(x))'.(sin^2(x)) + (cos^3(x)).(sin^2(x))'
Now use the chain rule for (cos^3(x))' and (sin^2(x))'
(cos^3(x))' = 3.cos^2(x).(-sin(x)) = -3.sin(x).cos^2(x)
(sin^2(x))' = 2.sin(x).cos(x)
So now then,
v' = (-3.sin(x).cos^2(x)).sin^2(x) + (cos^3(x)).(2.sin(x).cos(x))
v' = -3.sin^3(x).cos^2(x) + 2.sin(x).cos^4(x)
uv' = e^(ax){-3.sin^2(x) + 2.cos^2(x)}.sin(x).cos^2(x)
u'v = a.e^(ax).cos^3(x).sin^2(x)
y = uv' + u'v = e^(ax).sin(x).cos^2(x){a.sin(x).cos(x) + 2.cos^2(x) - 3.sin^2(x)}