differentiate y=x^√x

Rewrite the right hand side in terms of exponentials and logarithms, using the fact that

[tex]a=e^{\ln a}=\exp(\ln a)[/tex]

[tex]y=x^{\sqrt x}=\exp\left(\ln x^{\sqrt x}\right)=\exp\left(\sqrt x\ln x\right)[/tex]

Now differentiate both sides, applying the chain and product rules.

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\exp\left(\sqrt x\ln x\right)[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\exp\left(\sqrt x\ln x\right)\cdot\dfrac{\mathrm d}{\mathrm dx}\left(\sqrt x\ln x\right)[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dx}=x^{\sqrt x}\left(\dfrac{\mathrm d}{\mathrm dx}[\sqrt x]\ln x+\sqrt x\dfrac{\mathrm d}{\mathrm dx}[\ln x]\right)[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dx}=x^{\sqrt x}\left(\dfrac{\ln x}{2\sqrt x}+\dfrac{\sqrt x}x\right)[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac12x^{\sqrt x}\left(x^{-1/2}\ln x+2x^{-1/2}\right)[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac12x^{\sqrt x-1/2}\left(\ln x+2\right)[/tex]