How do you differentiate #f(x)=sec(1/sqrt(3x^2-4) ) # using the chain rule?

1 Answer
Jun 26, 2018

Answer:

#(dy)/(dx)=-(3x)/(3x^2-4)^(3/2)sec(1/sqrt(3x^2-4))tan(1/sqrt(3x^2-4))#

Explanation:

Let ,

#color(red)(y=secu # , #color(red)(u=1/v and v=sqrt(3x^2-4#

#=>(dy)/(du)=secutanu , # #(du)/(dv)=-1/v^2 and #

#(dv)/(dx)=1/(2sqrt(3x^2-4))d/(dx)(3x^2-4)=(6x)/(2sqrt(3x^2-4))=(3x)/sqrt(3x^2-4)#

Using chain rule:

#color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dv) .(dv)/(dx)#

#(dy)/(dx)=secutanu(-1/v^2)(3x)/sqrt(3x^2-4) ,where,u=1/v#

#(dy)/(dx)=sec(1/v)tan(1/v)(-1/v^2)(3x)/sqrt(3x^2-4) #

Putting , #v=sqrt(3x^2-4)#

#(dy)/(dx)#=#-sec(1/sqrt(3x^2-4))tan(1/sqrt(3x^2-4))1/(3x^2-4)*(3x)/sqrt(3x^2-4)#

#(dy)/(dx)=-(3x)/(3x^2-4)^(3/2)sec(1/sqrt(3x^2-4))tan(1/sqrt(3x^2-4))#