How do you find the derivative of $y = sin 2 x + cos 2 x + ln (e x)$ $y=sin2x+cos2x+ln(ex)$ ?

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How do you find the derivative of $y = sin 2 x + cos 2 x + ln (e x)$ $y=sin2x+cos2x+ln(ex)$ ?

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Answer 1 · 2015-03-06T20:38:48

Your expression is

$y = sin 2 x + cos 2 x + ln (e x)$ $y = sin2x + cos2x + ln(ex)$

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

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

$y = sin 2 x + cos 2 x + ln (x) + 1$ $y = sin2x + cos2x + ln(x) + 1$

Differentiating throughout with respect to x,

$\frac{d y}{d x} = \frac{d}{d x} (sin 2 x) + \frac{d}{d x} (cos 2 x) + \frac{d}{d x} (ln x) + \frac{d}{d x} 1$ $dy/dx = d/dx(sin2x) + d/dx(cos2x) + d/dx(lnx) + d/dx1$

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

$\frac{d y}{d x} = 2 cos 2 x - 2 sin 2 x + \frac{1}{x}$ $dy/dx = 2cos2x - 2sin2x + 1/x$

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

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How do you find the derivative of $y = sin 2 x + cos 2 x + ln (e x)$ $y=sin2x+cos2x+ln(ex)$ ?

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

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How do you find the derivative of $y = sin 2 x + cos 2 x + ln (e x)$ $y=sin2x+cos2x+ln(ex)$ ?