f^{'}(x) = e^{x}(tanx+sec^{2}x-x-1)
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Hey jayjay55531. Hint: Consider the relationship between tan^2(x) and (sec^2(x)).
Hmmm. Well sec^{2}(x) = 1+tan^{2}(x), right? so.... I'd end up with e^{x}(tanx+tan^{2}x-x), right? So then is that the most simplified it can be?
Thats pretty simplified: I wouldn't try and over-do it (it looks good).
Alright. Thanks so much for your help!
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