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
JN
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