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

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Answer 1 · 2016-11-09T05:24:58

#dy/dx=(secxtanx+sec^2x)/(1+(secx+tanx)^2)#

Rearranging:

#tany=secx+tanx#

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

#sec^2y*dy/dx=secxtanx+sec^2x#

Solving for the derivative:

#dy/dx=(secxtanx+sec^2x)/sec^2y#

Using the Pythagorean identity:

#dy/dx=(secxtanx+sec^2x)/(1+tan^2y)#

Using #tany=secx+tanx#:

#dy/dx=(secxtanx+sec^2x)/(1+(secx+tanx)^2)#