find the derivative of tan inverse ( sec x + tan x)

Question

anonymous · Answer

$$Let y = \tan ^{-1}(secx+tanx)$$
Differentiating both sides w.r.t x we find
$$\frac{ dy }{dx }=\frac{ d }{ dx }\tan ^{-1}(secx+tanx)$$
                         $$=\frac{ 1 }{ 1+ (secx+tanx)^{2} }\frac{ d }{ dx }(secx+tanx)$$
                         $$=\frac{ 1 }{ 1+ (secx+tanx)^{2} }(secx tanx+\sec ^{2}x)$$
$$=\frac{ secx(tanx+\sec x) }{ 1+ (secx+tanx)^{2} }$$
If this is the ans then stop here otherwise simplify it

hartnn · Answer

denominator = 1+(s+t)^2 = 1+s^2+2st+t^2
but, s^2 = 1+t^2
so, 2s^2+2st = 2s (s+t)
s(s+t) will get cancelled from numerator and denominator
only 1/2 remains...

anonymous · Answer

Yes , you are absolutely correct