Question

Find `d/dx x^(x^x)`?

Answer 1

`x^(x^x)*(x^xlnx){1+lnx+1/(xlnx)}`.

Explanation:

Suppose that, `y=x^(x^x)`.

`:. lny=(x^x)lnx`.

`:. ln(lny)=ln{(x^x)lnx}=ln(x^x)+ln(lnx)`,

`i.e., ln(lny)=xlnx+ln(lnx)`.

Diff.ing both sides w.r.t. `x`,

`d/dx{ln(lny)}=d/dx{xlnx+ln(lnx)}..................(star)`.

Here, by the Chain Rule,

`d/dx{ln(lny)}=1/lny*d/dx{lny}=1/lny*d/dy{lny}*dy/dx`.

`;. d/dx{ln(lny)}=1/lny*1/y*dy/dx=1/(ylny)*dy/dx`.

Utilising this in `(star)`, we get,

`1/(ylny)*dy/dx=d/dx{xlnx}+d/dx{ln(lnx)}`,

`=[x*d/dx{lnx}+(lnx)*d/dx{x}]+1/lnx*d/dx{lnx}`,

`=[x*1/x+(lnx)*1]+1/lnx*1/x`,

`rArr 1/(ylny)*dy/dx=1+lnx+1/(xlnx)`.

`:. dy/dx=(ylny){1+lnx+1/(xlnx)}`.

Here, sub.ing `y=x^(x^x), lny=x^xlnx`, we get,

`dy/dx=x^(x^x)*(x^xlnx){1+lnx+1/(xlnx)}`.

Answer 2

`d/dxx^(x^x)=x^(x^x)(x^x(ln^2x+lnx)+x^(x^x-1))`

Explanation:

In order to make this easier, we shall introduce a substitution:

Let `t=x^x` and so `x^(x^x)=x^t=exp(tlnx)`

So,

`d/dxx^(x^x)=d/dx exp(tlnx)=exp(tlnx)(d/dx(tlnx))=`

`x^t((dt)/dxlnx+t/x)=x^(x^x)(dt/dxlnx+x^(x^x)/x)`

Note: we found `d/dx(tlnx)` using the product rule

Now we need to find `dt/dx`

`t=x^x=exp(xlnx)` so `dt/dx=exp(xlnx)(d/dxxlnx)`

`=x^x(lnx+1)`

Putting this all together gives

`d/dxx^(x^x)=x^(x^x)(x^x(ln^2x+lnx)+x^(x^x-1))`

Answer 3

`= x^(x(1+x^(x-1))) (ln^2(x) + ln(x) + 1/x)`

Explanation:

`x^y = exp(y*ln(x))`

`=> (d/{dx}) x^y = (y*ln(x))' exp(y*ln(x))`

`= (y' ln(x) + y/x) x^y`

`=> (d/{dx}) x^x = (ln(x) + 1) x^x`

`=> (d/{dx}) x^(x^x) = ((ln(x)+1) x^x * ln(x) + x^x/x) x^(x^x)`

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

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

`= x^(x(1+x^(x-1))) (ln^2(x) + ln(x) + 1/x)`