Question

How do you find the derivative of `y = arctan [ tanx / (sqrt(2+tan^2(x))) ] + ln [tanx + sqrt(2+tan^2(x)) ]`?

Answer 1

`(dy)/(dx)=sqrt(2+tan^2x)`
The solution is to long to show all steps,with formulae and simplifiction.

Explanation:

Let ,`y=u+v ,where,`

`u=tan^-1[tanx/(sqrt(2+tan^2x))] and v=ln|tanx+sqrt(2+tan^2x)|`

• `(du)/(dx)=1/(1+(tanx/(sqrt(2+tan^2x)))^2)xxd/(dx)(tanx/(sqrt(2+tan^2x)))`
  `=cancel((2+tan^2x))/(2+tan^2x+tan^2x)xx(((sqrt(2+tan^2x))(sec^2x)-tanx((cancel(2)tanxsec^2x)/(cancel(2)sqrt(2+tan^2x))))/(cancel((2+tan^2x)))`
  `=((sec^2x(sqrt(2+tan^2x)-tan^2x/(sqrt(2+tan^2x))))/(2+2tan^2x))`
  `=(sec^2x(2+tan^2x-tan^2x))/(2(sec^2x)(sqrt(2+tan^2x)))=1/(sqrt(2+tan^2x))..........(I)`
• `(dv)/(dx)=1/(tanx+sqrt(2+tan^2x))(sec^2x+(2tanxsec^2x)/(2sqrt(2+tan^2x)))`
  `=sec^2x/(tanx+sqrt(2+tan^2x))(1+tanx/sqrt(2+tan^2x))`
  `=sec^2x/(tanx+sqrt(2+tan^2x))((sqrt(2+tan^2x)+tanx)/sqrt(2+tan^2x))`
  `=(1+tan^2x)/sqrt(2+tan^2x)..........(II)`
  Using `(I)and(II)`
  `(dy)/(dx)=(du)/(dx)+(dv)/(dx)=1/sqrt(2+tan^2x)+(1+tan^2x)/sqrt(2+tan^2x)=(2+tan^2x)/sqrt(2+tan^2x)=sqrt(2+tan^2x)`