# Find the derivative of the function. y=ln|sec x+tan x|

Find the derivative of the function. $y=\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|$
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Roosevelt Houghton
$y=\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|$
$\frac{dy}{dx}=\frac{d\left(\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|\right)}{dx}$
$\frac{dy}{dx}=\frac{1}{\left(\mathrm{sec}x+\mathrm{tan}x\right)}\cdot \frac{d\mathrm{sec}x+\mathrm{tan}x}{dx}$
$\frac{dy}{dx}=\frac{1}{\left(\mathrm{sec}x+\mathrm{tan}x\right)}\cdot \left(\mathrm{sec}x\mathrm{tan}x+{\mathrm{sec}}^{2}x\right)$
$\frac{dy}{dx}=\frac{1}{\left(\mathrm{sec}x+\mathrm{tan}x\right)}\cdot \mathrm{sec}x\left(\mathrm{tan}x+\mathrm{sec}x\right)$
$\frac{dy}{dx}=\frac{\left(\mathrm{sec}x+\mathrm{tan}x\right)}{\left(\mathrm{sec}x+\mathrm{tan}x\right)}\cdot \mathrm{sec}x$
$\frac{dy}{dx}=\mathrm{sec}x$
Result
$\frac{dy}{dx}=\mathrm{sec}x$