How do you differentiate # f(x)=sqrt(ln(1/(xe^x))# using the chain rule.?

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Answer 1 · 2016-11-15T16:31:08

#(-1/2)((x+1)/(xf(x)))#

#e^x > 0#. So, for ln to be real, #xe^x > 0 to x > 0#

Also, for sqrt to be real, ln > 0, .

So, #xe^x > 1#.

By a numerical method,

x < 0.567, nearly.

Thus, f(x) is defined for #x in ( 0, 0.567 )

Use #ln(1/(xe^x))=-ln(xe^x)=-lnx - ln(e^x=-ln x - x#

For this x,

#f'=((-ln x - x)^(1/2))'#

#=(1/2)((-ln x - x)^(-1/2))(-x-ln x)'#

#=(-1/2)((-ln x - x)^(-1/2))(-1-1/x)#

#=(-1/2)((x+1)/(xf(x)))#