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)#