Question

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

Answer 1

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.