How do you find the equation of the tangent line to the graph of #y=ln(x*e^x)# at x=1?

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Answer 1 · 2016-11-22T02:14:29

Take the derivative of your function, then set #x=1#

#y' = (e^x+xe^x)/(xe^x)=(x+1)/x#

Plug in #x=1#

#((1)+1)/(1) = 2#

This is the slope of your tangent line.

Now we need a y-coordinate for our equation - lets go get it.

#y=ln(1*e^1) = lne = 1#

Use point-slope formula or whatever you're comfortable with:

#y-y_1 = m(x-x_1)#

#y-1=2(x-1)#

#y=2x-1 -> #Your tangent line equation.