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