How do you find the derivative of $y = arctan (sec x + tan x)$ $y=arctan(secx + tanx)$ ?

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Answer 1 · 2016-11-09T05:24:58

$\frac{d y}{d x} = \frac{sec x tan x + {sec}^{2} x}{1 + {(sec x + tan x)}^{2}}$ $dy/dx=(secxtanx+sec^2x)/(1+(secx+tanx)^2)$

Rearranging:

$tan y = sec x + tan x$ $tany=secx+tanx$

Differentiating both sides, and recalling to use the chain rule on the left:

${sec}^{2} y \cdot \frac{d y}{d x} = sec x tan x + {sec}^{2} x$ $sec^2y*dy/dx=secxtanx+sec^2x$

Solving for the derivative:

$\frac{d y}{d x} = \frac{sec x tan x + {sec}^{2} x}{{sec}^{2} y}$ $dy/dx=(secxtanx+sec^2x)/sec^2y$

Using the Pythagorean identity:

$\frac{d y}{d x} = \frac{sec x tan x + {sec}^{2} x}{1 + {tan}^{2} y}$ $dy/dx=(secxtanx+sec^2x)/(1+tan^2y)$

Using $tan y = sec x + tan x$ $tany=secx+tanx$ :

$\frac{d y}{d x} = \frac{sec x tan x + {sec}^{2} x}{1 + {(sec x + tan x)}^{2}}$ $dy/dx=(secxtanx+sec^2x)/(1+(secx+tanx)^2)$

How do you find the derivative of y=arctan(secx + tanx)y=arctan(secx+tanx)y=arctan(secx + tanx)?