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For #f(x)=e^(xlnx)# what is the equation of the tangent line at #x=1#?

Answer 1

#y=x#

#f(1)=e^(1ln1)=1#

#f'(x)=e^(xlnx)*(x*1/x+lnx)=(1+lnx)*e^(xlnx)#

#f'(1)=1#

Therefore tangent line is straight line so of form #y=mx+c#, with gradient #m=1#. Substituting point #(1,1)# into equation of tangent yields #1=(1)(1)+c=>c=0#.

Hence equation of required tangent is #y=x#.

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Answer 2

Paisley Johnson

To find the equation of the tangent line at x=1 for the function f(x)=e^(xlnx), we need to find the derivative of the function and evaluate it at x=1. The derivative of f(x) can be found using the chain rule and product rule. After finding the derivative, we substitute x=1 into the derivative to find the slope of the tangent line. Finally, we use the point-slope form of a line to write the equation of the tangent line.

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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.