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

Mar 6, 2015

$y = \sin 2 x + \cos 2 x + \ln \left(e x\right)$

The last term can be written as $\ln \left(e x\right) = \ln \left(e\right) + \ln \left(x\right)$

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

$y = \sin 2 x + \cos 2 x + \ln \left(x\right) + 1$

Differentiating throughout with respect to x,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\sin 2 x\right) + \frac{d}{\mathrm{dx}} \left(\cos 2 x\right) + \frac{d}{\mathrm{dx}} \left(\ln x\right) + \frac{d}{\mathrm{dx}} 1$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos 2 x - 2 \sin 2 x + \frac{1}{x}$

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