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

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Answer 1 · 2017-07-19T14:04:18

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

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

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

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

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

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

or

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

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