Respuesta :

LammettHash
[tex]y=x^{x^2}[/tex]
[tex]\ln y=\ln x^{x^2}[/tex]
[tex]\ln y=x^2\ln x[/tex]

Now differentiating with respect to [tex]x[/tex] gives

[tex]\dfrac{y'}y=2x\ln x+\dfrac{x^2}x[/tex]
[tex]\dfrac{y'}y=2x\ln x+x[/tex]
[tex]\dfrac{y'}y=x(\ln x^2+1)[/tex]
[tex]y'=yx(\ln x^2+1)[/tex]
[tex]y'=x^{x^2}x(\ln x^2+1)[/tex]
[tex]y'=x^{x^2+1}(\ln x^2+1)[/tex]

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