Can someone explain Implicit Differentiation and Related Rates in Calculus?

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SUMMARY

The discussion centers on the concepts of Implicit Differentiation and Related Rates in Calculus, specifically addressing the challenges faced by students in understanding these topics. A key example provided is the equation xy = 1, which illustrates the application of the product rule in differentiation. The explanation emphasizes treating y as a function of x and differentiating accordingly, leading to the equation y(x) + x y'(x) = 1. Additionally, a resource from the University of Tennessee is recommended for further clarification.

PREREQUISITES
  • Understanding of basic differentiation rules, including the product rule.
  • Familiarity with implicit functions and their derivatives.
  • Knowledge of algebraic manipulation to solve for unknowns.
  • Basic calculus concepts, particularly related rates.
NEXT STEPS
  • Study the application of the product rule in implicit differentiation.
  • Explore related rates problems and their solutions in calculus.
  • Review resources on implicit differentiation, such as the one from the University of Tennessee.
  • Practice solving equations involving multiple variables and their derivatives.
USEFUL FOR

Students enrolled in Calculus courses, educators teaching calculus concepts, and anyone seeking to improve their understanding of implicit differentiation and related rates.

Azrioch
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Hi. I'm taking a Calculus course right now and I simply cannot understand Implicit Differentiation or the Related Rate problems. My textbook does not do a good job explaining it. It is a very accelerated class and I cannot get it and I need to know it in two days for a mid term.

I just don't understand the concept.. or well any of it.

Could someone explain it to me?

Thank you in advance.
 
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Forget you're doing something "fancy" and just apply the rules of differentiation... remembering that if you're differentiating with respect to, say, x, the other variables are functions of x.

e.g.

xy = 1

rememebr that y is a function of x, so write as:

x y(x) = 1

then differentiate both sides using the product rule

y(x) + x y'(x) = 1

and now if you want something interesting like y'(1), you just have an ordinary algebra problem... you have two unknowns, y(1) and y'(1), and two equations.
 
Hi!

here is a good site for example to help you. http://archives.math.utk.edu/visual.calculus/3/implicit.7/ I find it easier to use the power rule for finding Dx after you have all variables to one side = 0.
Dx :wink:
 

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