Question

Differentiate `log [x+2+sqrt(x^2+4x+1)]`

Answer 1

Let y = `log [x+2+sqrt(x^2+4x+1)]`

Differentiate it with respect to x we get,

`(dy)/(dx)=d/(dx)log[x+2+sqrt(x^2+4x+1)]`

`=1/([x+2+sqrt(x^4+4x+1)])d/(dx)[x+2+(x^2+4x+1)^(1/2)]`      [Using chain rule]

`=1/([x+2+sqrt(x^4+4x+1)])xx[1+0+1/2(x^2+4x+1)^(-1/2)d/(dx)(x^2+4x+1)]`

`=(1+(2x+4)/(2(sqrt(x^2+4x+1))))/([x+2+sqrt(x^4+4x+1)])`

`=(sqrt(x^2+4x+1)+x+2)/([x+2+sqrt(x^2+4x+1)]xxsqrt(x^2+4x+1))`

`=1/sqrt(x^2+4x+1)`

So, `d/(dx)log[x+2+sqrt(x^2+4x+1)]=1/sqrt(x^2+4x+1)`