How Can We Calculate the Derivative of x^x^x?

[mubeenahm]

let y=x^x^x
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]

 

[HallsofIvy]

let y=x^x^x

I assume you mean x^(x^x) since (x^x)^x= x^(x^2)

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]

Looks good to me.