Question

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

Answer 1

\[Let y = \tan ^{-1}(secx+tanx)\]
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

Answer 2

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...

Answer 3

Yes , you are absolutely correct