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

3 Answers
May 24, 2018

# 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)}#.

May 24, 2018

#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))#

#= 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)#