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How do you use logarithmic differentiation for yxex? - Answers

I take this to be y = xex.Proceeding formally (ie, without regard to restrictions on the domain of ln x):Take natural logarithms of 'both sides' of above equation: ln y = ln x + xImplicit differentiation: y'/y = 1/x + 1Multiply both sides by y: y' = y ( 1/x + 1 )Replace y by its definition as a function: y' = xex ( 1/x + 1 ).



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How do you use logarithmic differentiation for yxex? - Answers

https://math.answers.com/calculus/How_do_you_use_logarithmic_differentiation_for_yxex

I take this to be y = xex.Proceeding formally (ie, without regard to restrictions on the domain of ln x):Take natural logarithms of 'both sides' of above equation: ln y = ln x + xImplicit differentiation: y'/y = 1/x + 1Multiply both sides by y: y' = y ( 1/x + 1 )Replace y by its definition as a function: y' = xex ( 1/x + 1 ).



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https://math.answers.com/calculus/How_do_you_use_logarithmic_differentiation_for_yxex

How do you use logarithmic differentiation for yxex? - Answers

I take this to be y = xex.Proceeding formally (ie, without regard to restrictions on the domain of ln x):Take natural logarithms of 'both sides' of above equation: ln y = ln x + xImplicit differentiation: y'/y = 1/x + 1Multiply both sides by y: y' = y ( 1/x + 1 )Replace y by its definition as a function: y' = xex ( 1/x + 1 ).

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      I take this to be y = xex.Proceeding formally (ie, without regard to restrictions on the domain of ln x):Take natural logarithms of 'both sides' of above equation: ln y = ln x + xImplicit differentiation: y'/y = 1/x + 1Multiply both sides by y: y' = y ( 1/x + 1 )Replace y by its definition as a function: y' = xex ( 1/x + 1 ).

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