Question

What is the Maclaurin series for `e^xsin(x)`?

Answer 1

The Maclaurin series for `f(x)` is

`sum_(n=0)^oo (f^('n)(0))/(n!) * (x)^n`

I assume the major difficulty here is evaluating
`f^('n)(0)` for successive terms of the series

Before getting into individual derivatives, let's first note:
`e^0 = 1`
`cos(0) = 1`
`sin(0) = 0`
`(d e^x)/(dx) = e^x`
`(d sin(x)) = cos(x)` and
`(d cos(x)) = - sin(x)`

If `f(x) = e^x sin(x)`, then

`f'(x) = (d e^x) / (dx) * sin(x) + e^x*(d sin(x))/(dx)`

`=e^x sin(x) + e^x cos(x)`

so `f'(0) = 1`

`f''(x) = ((d (e^x sin(x)))/(dx)) + ((d (e^x cos(x)))/(dx))`
`= (e^x sin(x) + e^x cos(x)) + (e^x cos(x) + e^x (-sin(x)) )`

`=2 e^x cos(x)`

and `f''(0) = 2`

continuing on in the same manner we can evaluate subsequent derivatives.

Simply insert these values into the definition of the Maclaurin series.