Question

D\dx(e*x+logx)\sin3x

Answer 1

Let t= (e^x + log x) / (sin3x)

dt / dx = d [ (e^x + log x) / (sin3x) ] / dx

Using the quotient rule, i.e (u/v)' = (vu' - v'u) / v^2

dt / dx = [ sin3x (e^x + 1/x)  - (e^x + logx).3.cos3x ] / sin^2 (3x)

           = [ sin3x (e^x + 1/x)  - 3(e^x + logx).cos3x ] / sin^2 (3x)

Note:

d(logx)dx = 1/x

d(e^x)/dx = e^x