What is the derivative of #y=ln(sec(x)+tan(x))#?

2 Answers

Answer: #y'=sec(x)#

Full explanation:

Suppose, #y=ln(f(x))#

Using chain rule, #y'=1/f(x)*f'(x)#

Similarly, if we follow for the problem, then

#y'=1/(sec(x)+tan(x))*(sec(x)+tan(x))'#

#y'=1/(sec(x)+tan(x))*(sec(x)tan(x)+sec^2(x)) #

#y'=1/(sec(x)+tan(x))*sec(x)(sec(x)+tan(x))#

#y'=sec(x)#

Apr 18, 2015

Will give you a personal video explanation of how it's done...

Learn how to differentiate y=ln(secx+tanx) in this video

Alternatively, you can use these workings...

#ln(secx+tanx)=y#

#e^y=secx+tanx#

#e^y*(dy)/(dx)=secxtanx+sec^2x#

#e^y*(dy)/(dx)=secx(secx+tanx)#

#(dy)/(dx)=(secx(secx+tanx))/e^y#

#(dy)/(dx)=(secx(secx+tanx))/((secx+tanx))#

#(dy)/(dx)=secx#