What is the derivative of $e^3x cos2x$ ?

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Answer 1 · 2015-08-06T19:14:12

$y^' = e^3 * [cos(2x) - 2x * sin(2x)]$

To differentiate this function, you can use the product rule and the chain rule. Keep in mind that you have

$d/dx(cosx) = -sinx$

So, according to the product rule, you can differentiate a function that takes the form

$y = f(x) * g(x)$

by using the formula

$color(blue)(d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))$

In your case, $f(x) = x$ and $g(x) = cos(2x)$ , which means that you can write

$d/dx(y) = e^3 ([d/dx(x)] * cos(2x) + x * underbrace(d/dx(cos(2x)))_(color(red)("use chain rule!")))$

$y^' = e^3 * [1 * cos(2x) + x * (-sin(2x) * d/dx(2x))]$

$y^' = e^3 * [cos(2x) - x * sin(2x) * 2]$

$y^' = color(green)(e^3 * [cos(2x) - 2x * sin(2x)])$

What is the derivative of e^3x cos2xe^3x cos2x?