Question

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

Answer 1

`(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))`