Derivative of square root of tan 2x?

2 Answers
Alan N.
May 6, 2018

#sec^2(2x)/sqrt(tan2x)#

Explanation:

#f(x) = sqrt(tan2x)#

#= (tan2x)^(1/2)#

Apply chain rule

#f'(x) = 1/2 (tan2x)^(-1/2) * d/dx(tan2x)#

Apply chain rule again

#f'(x) = 1/(2sqrt(tan2x)) * sec^2(2x) * d/dx 2x#

Apply power rule

#f'(x) = 1/(2sqrt(tan2x)) * sec^2(2x) * 2#

#= 1/(cancel(2)sqrt(tan2x)) * sec^2(2x) * cancel2#

#= sec^2(2x)/sqrt(tan2x)#

maganbhai P.
May 6, 2018

#y=sqrt(tan2x)=>(dy)/(dx)=1/(2sqrt(tan2x))xxd/(dx)(tan2x)#
#=>1/(2sqrt(tan2x))xxsec^2(2x)d/(dx)(2x)=sec^2(2x)/(2sqrttan(2x))xx2#
#=>(dy)/(dx)=sec^2(2x)/sqrt(tan2x)#

Explanation:

Here,

#y=sqrt(tan2x)#

Let, #y=sqrtu#, where , #u=tanv and v=2x#

#=>(dy)/(du)=1/(2sqrtu) , (du)/(dv)=sec^2v and (dv)/(dx)=2#

#"Using " color(blue)"Chain Rule"#, we get

#color(blue)((dy)/(dx)=(dy)/(du)xx(du)/(dv)xx(dv)/(dx)#

#=>(dy)/(dx)=1/(2sqrtu)xxsec^2vxx2=sec^2v/sqrtu#

Subst. #sqrtu=y=sqrt(tan2x) and v=2x,#we get

#(dy)/(dx)=sec^2(2x)/sqrt(tan2x)#

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