Remembering the derivative of arctanx here is very helpful. If you don't, it could get a bit frustrating with implicit differentiation. I'll do it both ways to show you what I mean.
d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))
Thus, we have:
y = arctan(secx + tanx)
(dy)/(dx) = 1/(1+(secx + tanx)^2)(secxtanx + sec^2x)
= (secxtanx + sec^2x)/(1+(secx + tanx)^2)
= (secxtanx + sec^2x)/(1+sec^2x + 2secxtanx + tan^2x)
= (sec^2x + secxtanx)/(2sec^2x + 2secxtanx)
= color(blue)(1/2)
(Haha, nice. A derivative that doesn't even integrate back into the original function without some special manipulation.)
And now the other way.
tany = secx + tanx
sec^2y * (dy)/(dx) = secxtanx + sec^2x
(1 + tan^2color(green)(y)) * (dy)/(dx) = secxtanx + sec^2x
Note that you have to use this identity, otherwise you'll get sec^2(arctan(secx + tanx)), which is not easy to work with.
[1 + (tan[color(green)(arctan(secx + tanx))])^2] (dy)/(dx) = secxtanx + sec^2x
[1 + (secx + tanx)^2] (dy)/(dx) = secxtanx + sec^2x
(dy)/(dx) = (secxtanx + sec^2x)/(1 + (secx + tanx)^2)
which, from above, is:
color(blue)(= 1/2)
