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

1 Answer
Mar 25, 2018

Answer:

#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)#