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How do you differentiate # f(x)=ln(sqrt(e^x-x))# using the chain rule.?

Answer 1

Samantha Adkison

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

#f(x)=ln(sqrt(e^x-x))#

#f'(x) = 1/(sqrt(e^x-x)) * d/dx(sqrt(e^x-x))# (Standard differential and chain rule)

#= 1/(sqrt(e^x-x)) * 1/2(e^x-x)^(-1/2) * d/dx(e^x-x)# (Power rule and chain rule)

#=1/(sqrt(e^x-x)) * 1/(2 (sqrt(e^x-x))) *(e^x-1)# (Standard differential and power rule)

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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