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

1 Answer

A. S. Adikesavan

Jul 10, 2016

#f'=(-1/2)(1+x)e^xf^3, x >=0.5672#, nearly.

Explanation:

The function is differentiable for #xe^x > 1# to make #sqrt ln(xe^x)#

value real. Numerical, method(s) give the corresponding value for x

to be #>=0.5672#, nearly..

#f'=(-1/2)(1/(ln(xe^x))^(3/2))(xe^x)'#

#=(-1/2)(1+x)e^xf^3#

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