How to find the derivative?

ln sqrt x^4+x^2+1

2 Answers
Nallasivam V
Feb 8, 2018

#dy/dx=( (4x^3+2x))/(2sqrt(x^4+x^2+1))#

Explanation:

Given -

#y=sqrt(x^4+x^2+1#

#y=(x^4+x^2+1)^(1/2)#

#dy/dx=1/2(x^4+x^2+1)^(1-1/2)(4x^3+2x)#

#dy/dx=1/2(x^4+x^2+1)^(-1/2)(4x^3+2x)#

#dy/dx=( 1xx (4x^3+2x))/(2sqrt(x^4+x^2+1))#

#dy/dx=( (4x^3+2x))/(2sqrt(x^4+x^2+1))#

Nallasivam V
Feb 8, 2018

#dy/dx=( (4x^3+2x))/(2sqrt(x^4+x^2+1))#

Explanation:

Given -

#y=sqrt(x^4+x^2+1#

#y=(x^4+x^2+1)^(1/2)#

#dy/dx=1/2(x^4+x^2+1)^(1-1/2)(4x^3+2x)#

#dy/dx=1/2(x^4+x^2+1)^(-1/2)(4x^3+2x)#

#dy/dx=( 1xx (4x^3+2x))/(2sqrt(x^4+x^2+1))#

#dy/dx=( (4x^3+2x))/(2sqrt(x^4+x^2+1))#

Related questions
Impact of this question
339 views around the world
You can reuse this answer
Creative Commons License