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

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Answer 1 · 2015-03-06T20:38:48

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.

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