Question

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

Answer 1

`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`