Question

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

Answer 1

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)`

Answer 2

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`