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find the equation of the tangent line to the graph of y= f(x) = ln (x)/x at the point (1,0)
1 answer
f'(x) = (1-lnx)/x^2
f'(1) = 1
So, now you have a point and a slope. The line is
y-0 = 1(x-1)
y=x-1
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