Question

How do you differentiate `f(x)=sqrt(e^((lnx-2)^2` using the chain rule.?

Answer 1

`f^'(x)=[(lnx-2)*e^((lnx-2)^2)]/(x*sqrt(e^((lnx-2)^2)))=1/x(lnx-2)sqrt(e^((lnx-2)^2)`

Explanation:

Here,
`f(x)=y=sqrt(e^((lnx-2)^2)`

Let us substitute,

`y=sqrtu, u=e^w and w=(lnx-2)^2`

`=>(dy)/(du)=1/(2sqrtu), (du)/(dw)=e^w and (dw)/(dx)=2(lnx-2)^1*1/x`

Using Chain Rule ,

`color(red)((dy)/(dx)=(dy)/(du)xx(du)/(dw)xx(dw)/(dx)`

`=>(dy)/(dx)=1/(cancel2sqrtu)xxe^wxx(cancel2(lnx-2))/x`

Substituting value of `u,v and w`

`(dy)/(dx)=1/sqrt(e^w)xxe^((lnx-2)^2)xx(lnx-2)/x`

`(dy)/(dx)=[(lnx-2)*e^((lnx-2)^2)]/(x*sqrt(e^((lnx-2)^2)`

`=>f^'(x)=1/x(lnx-2)sqrt(e^((lnx-2)^2)`