# What is derivative of `sqrt(x^x)`?

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### 1 Answer

Let, `y = sqrt(x^x)=x^(x^(1/2))`

`rArr y= x^(1/2*x)` [since `u^(v^z)=u^(v*z)` ]

Taking log of both sides,

`lny = ln(x^(1/2*x)) =(1/2)xlnx` [since `ln(u^v)=vlnu` ]

For differentiation, the left hand side of the equation requires chain rule whereas the right hand side requires product rule. Thus,

`1/y*(dy)/(dx)=1/2[lnx*1+x*(1/x)]`

`rArr 1/y*(dy)/(dx)=1/2[(lnx+1)]`

Multiplying both sides by y, to eliminate it from the L.H.S., we get

`(dy)/(dx)= 1/2[y(lnx+1)]=1/2*sqrt(x^x)(lnx+1)`

Therefore, the derivative of `sqrt(x^x)` is `1/2*sqrt(x^x)(lnx+1)` (assuming, x>0)

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