Derivatives of exponential functions

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by y=e^x that is perpendicular to the line 3x+y=1. The slope of the tangent line is determined to be 1/3, which is the negative reciprocal of the slope of the given line. The derivative of e^x is e^x itself, leading to the equation 1/3=e^x. The final solution is expressed as x-3y+(1+ln3)=0, utilizing the natural logarithm to solve for x.

PREREQUISITES
  • Understanding of exponential functions, specifically y=e^x
  • Knowledge of derivatives and their applications
  • Familiarity with the concept of perpendicular lines and slopes
  • Basic skills in solving logarithmic equations
NEXT STEPS
  • Study the properties of exponential functions and their derivatives
  • Learn about the concept of tangent lines in calculus
  • Explore the method of finding slopes of perpendicular lines
  • Practice solving logarithmic equations, particularly involving natural logarithms
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and tangent lines, as well as educators looking for examples of exponential function applications.

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Homework Statement



find the equation of the tangent to the curve defined by y=e^x that is perpendicular to the line defined by 3x+y=1

Homework Equations



m= 1/3 (negative reciprocal of y=-3x+1)
derivative of e^x=e^x

The Attempt at a Solution



i keep getting stuck at 1/3=e^x ( i know there's a way to solve for x i just can't remember)

the answer is x-3y+(1+ln3)=0
 
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ln(e^x) = x. You can take the natural log of both sides of an equation.
 

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