# What is the derivative of #y=arctan(secx + tanx)#?

Thus, we have:

(Haha, nice. A derivative that doesn't even integrate back into the original function without some special manipulation.)

And now the other way.

which, from above, is:

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To find the derivative of ( y = \arctan(\sec x + \tan x) ), we'll first need to use the chain rule. The derivative of ( \arctan(u) ) with respect to ( u ) is ( \frac{1}{1 + u^2} ), and then we'll multiply by the derivative of the inside function.

So, let's differentiate ( \sec x + \tan x ) with respect to ( x ). The derivative of ( \sec x ) is ( \sec x \tan x ), and the derivative of ( \tan x ) is ( \sec^2 x ).

Putting it all together:

[ \begin{align*} \frac{d}{dx} \left( \arctan(\sec x + \tan x) \right) &= \frac{1}{1 + (\sec x + \tan x)^2} \cdot \left( \frac{d}{dx} (\sec x + \tan x) \right) \ &= \frac{1}{1 + (\sec x + \tan x)^2} \cdot (\sec x \tan x + \sec^2 x) \end{align*} ]

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