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put natural logarithm ln on both sides

ln y=ln x^x^x

we know that ln a^b= b*(ln a)

let x^x=b

therefore: ln y= (x^x)*(ln x)

now,taking the derivative of both sides

derivative of ln y=y`/y

and x^x=u,ln x= v.........using the u*v rule we get=u`v+v`u(where ` is the derivative)

so we have:

y`/y=(x^x)`* (ln x)+(ln x)` * (x^x) [the derivative of x^x is =x^x(ln x+1) and derivative of ln x=1/x]

so, y`/y= x^x(ln x+1)*(ln x)+(1/x)*(x^x)

y`/y=x^x(ln^2 x+ln x)+(1/x)*(x^x)

taking x^x common on the right hand side

y`/y=x^x[ln^2 x +ln x +1/x]

now the y dividing y` goes to the right hand side and multiplies

y`=y*(x^x)[ln^2 x +ln x+1/x]

we know that y=x^x^x

therefore y`=(x^x^x)*(x^x)[ln^2 x +ln x+1/x]

since the bases are same ,powers should be added

so the final answer becomes

y`=x^[(x^x)+x][ln^2 x +ln x+1/x]

OR

d(y)/dx=x^[(x^x)+x][ln^2 x +ln x+1/x]

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# The derivative of x^x^x

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