Given that y = ((4x+3)^5)(sin2x), find dy/dx

First of all, we have to use the product rule, since two things are multiplied together.  The product rule states that d/dx (u*v) = vu' + uv'

Let u = (4x+3)⁵  v = sin(2x)

Now, to find  u' and v' we have to find du/dx and dv/dx. As we see, we need to use the chain rule to find du/dx

u' = 20(4x+3)⁴   v' = 2cos(2x)

Finally, dy/dx = vu' + uv' = 20sin(2x)(4x+3)⁴ + 2cos(2x)(4x+3)⁵