How to differentiate this by using chain rule?

#y=ln(sqrt(x^2+5))#

1 Answer
maganbhai P.
Mar 25, 2018

#(dy)/(dx)=x/(x^2+5)#

Explanation:

Here,

#y=lnsqrt(x^2+5)#

#=>(dy)/(dx)=1/sqrt(x^2+5)d/(dx)(sqrt(x^2+5))#

#=>(dy)/(dx)=1/sqrt(x^2+5)xx1/(2sqrt(x^2+5))d/(dx)(x^2+5)#

#=>(dy)/(dx)=1/sqrt(x^2+5)xx1/(cancel2sqrt(x^2+5))(cancel2x)#

#=>(dy)/(dx)=x/(x^2+5)#

OR take,

#y=lnu, and u=sqrt(x^2+5)#

#=>(dy)/(du)=1/u, and (du)/(dx)=1/(2sqrt(x^2+5))*2x=x/sqrt(x^2+5)#

Using chain rule,

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

#=1/uxxx/(sqrt(x^2+5)#

#=1/sqrt(x^2+5)xxx/sqrt(x^2+5)#

#=x/(x^2+5)#

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