If y=ln(x^2) find: The equation to the tangent line where x=e The point where the slope of the tangent line is 2/3 HELP.
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Originally Posted by puravida12 If y=ln(x^2) find: The equation to the tangent line where x=e The point where the slope of the tangent line is 2/3. $\displaystyle y = \ln \left( {x^2 } \right) = 2\ln (x)\, \Rightarrow \,y' = \frac{2} {x}$
Thanks, but what exactly does it mean when it says "where x=e"? and how do I find the point when given "2/3" as a tangent line?
Originally Posted by puravida12 what exactly does it mean when it says "where x=e"? Sorry but one does not expect that sort of question from someone in a calculus course. $\displaystyle y = \ln \left( {x^2 } \right),\,x = e\, \Rightarrow \,y = 2$
...what is that supposed to mean? thank you, sorry, I was confused.
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