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find the equation of the tangent line y = ln(x) to the graph at x=e
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start by finding the slope of the tangent line by finding y' at x = e
the derivative is 1/x
right, and what's its value when x = e?
1/e
1/e
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yeah from here you can use the point slope formula.\[y-y_1=m(x-x_1)\] We found m = 1/3 and were given \(x_1\)= e. Plug in e to the original equation to get \(y_1\)
one moment while i solve for this please...
y-lne=1/3(x-e) ?
sorry I made a typo. 1/3 is supposed to be 1/e. and you can simplify \(\ln e\) to 1
oh right! so: y-1=1/e(x-e)
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why is the slope 1/e
because the derivative is the slope of the tangent line. Taking the derivative then plugging in e gave the slope
ah! i get it! thank you so much
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