Find equations of the tangent lines to the curve

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SUMMARY

The discussion focuses on finding the equations of the tangent lines to the curve defined by the function y=(lnx)/x at the specific points (1,0) and (e,1/e). The derivative of the function is established as dy/dx = (1 - lnx)/x^2, which provides the gradient of the tangent line at any point 'x'. To find the equations of the tangent lines, one must evaluate this derivative at the given points and use the point-slope form of a line.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with the natural logarithm function (ln)
  • Knowledge of the point-slope form of a linear equation
  • Ability to graph functions and their derivatives
NEXT STEPS
  • Calculate the tangent line equations using the point-slope form with the gradients found at (1,0) and (e,1/e)
  • Graph the function y=(lnx)/x along with its tangent lines for visual confirmation
  • Explore the implications of the derivative in relation to curve behavior
  • Study higher-order derivatives to understand concavity and inflection points of the function
USEFUL FOR

Students and educators in calculus, mathematicians interested in curve analysis, and anyone looking to deepen their understanding of derivatives and tangent lines.

balla123
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Find equations of the tangent lines to the curve y=(lnx)/x
at the points (1,0) and (e,1/e) . Illustrate by graphing the
curve and its tangent lines.


i found that the derivative is 1-lnx/x^2

what do i do next? thanks
 
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I did not check your derivative, but you have dy/dx = (1-lnx)/x2

to get the gradient at any point 'x'. Just put that value of 'x' in the expression for 'dy/dx' and you will get the gradient of the tangent at that point.
 

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