How do you differentiate #e^(4x)*sin(4x)#?

1 Answer
Jul 19, 2017

#d/(dx)[e^(4x)sinx] = color(blue)(e^(4x)cosx + 4e^(4x)sinx#

Explanation:

We're asked to find the derivative

#d/(dx) [e^(4x)sinx]#

We can first use the product rule, which states

#d/(dx)[uv] = v(du)/(dx) u(dv)/(dx)#

where

  • #u = e^(4x)#

  • #v = sinx#:

#= e^(4x)d/(dx)[sinx] + sinxd/(dx)[e^(4x)]#

The derivative of #sinx# is #cosx#:

#= e^(4x)cosx + sinxd/(dx)[e^(4x)]#

Use the chain rule to differentiate the #e^(4x)# term:

#d/(dx) [e^(4x)] = d/(du)[e^u] (du)/(dx)#

where

  • #u = 4x#

  • #d/(du)[e^u] = e^u#:

#= e^(4x)cosx + (sinx)(e^(4x))d/(dx)[4x]#

#= e^(4x)cosx + (sinx)(e^(4x))(4)#

or

#= color(blue)(e^(4x)cosx + 4e^(4x)sinx#