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

Question

7505 views around the world

You can reuse this answer
Creative Commons License

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)])#