Question

differentiation y=log(secx/2+tanx/2)

Answer 1

$y = log(sec(x/2) + tan(x/2))$
put 
$u = sec(x/2) + tan(x/2)$
$\frac{du}{dx} = sec (x/2) tan (x/2)( \frac{1}{2} ) + sec^2 (x/2)( \frac{1}{2} )$
then
[tex] \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{u} \frac{du}{dx} \\ \frac{dy}{dx} = \frac{sec (x/2) tan (x/2)( \frac{1}{2} ) + sec^2 (x/2)( \frac{1}{2} ) }{sec(x/2) + tan(x/2)} [/tex]
$\frac{dy}{dx} = \frac{sec(x/2)}{2}$

Answer 2

Answer:

Step-by-step explanation: