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

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Answer 1 · 2018-03-29T10:04:46

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

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 $(I)and(II)$
$(dy)/(dx)=(du)/(dx)+(dv)/(dx)=1/sqrt(2+tan^2x)+(1+tan^2x)/sqrt(2+tan^2x)=(2+tan^2x)/sqrt(2+tan^2x)=sqrt(2+tan^2x)$

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