How do you find the derivative of #y=sin2x+cos2x+ln(ex)#?

1 Answer
Mar 6, 2015

Your expression is

#y = sin2x + cos2x + ln(ex)#

The last term can be written as #ln(ex) = ln(e) + ln(x)#

#ln(e)# equals 1. Therefore, the original expression becomes,

#y = sin2x + cos2x + ln(x) + 1#

Differentiating throughout with respect to x,

#dy/dx = d/dx(sin2x) + d/dx(cos2x) + d/dx(lnx) + d/dx1#

You can now apply the chain rule of differentiation to the first two terms on the right hand side of the equation, to get this

#dy/dx = 2cos2x - 2sin2x + 1/x#

The last term is a constant, so its derivative is 0.