# `y = sqrt(x)^x` Use logarithmic differentiation to find the derivative of the function.

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Expert Answers

embizze | Certified Educator

Differentiate ` y=sqrt(x)^x ` :

Rewrite the right hand side:

`y=x^(x/2) ` (Note that `sqrt(x)=x^(1/2) ` so `sqrt(x)^x=(x^(1/2))^x=x^(x/2) ` )

Take the natural logarithm of both sides:

`lny=lnx^(x/2) `

Use a property of logs: Note that `loga^b=bloga `

`lny=x/2lnx `

Differentiate both sides; use the product property on the RHS:

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

`(dy)/(dx)(1/y)=1/2lnx+1/2 `

`(dy)/(dx)=y/2(lnx+1) `

Substituting for y we get:

`y'=1/2 x^(x/2)(lnx+1) `