Simplify the Differential Quotient | f(x) = 5x^2

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SUMMARY

The discussion focuses on simplifying the difference quotient for the function f(x) = 5x². Participants clarify that the limit as h approaches zero is necessary to transition from the difference quotient to the derivative. The correct approach involves calculating f(x+h) as 5(x+h)², then subtracting f(x) and dividing the result by h. This process leads to the simplification of the difference quotient.

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  • Understanding of basic calculus concepts, specifically limits.
  • Familiarity with polynomial functions and their properties.
  • Knowledge of the difference quotient and its role in finding derivatives.
  • Ability to perform algebraic manipulations, including expanding binomials.
NEXT STEPS
  • Study the concept of limits in calculus, particularly the limit definition of a derivative.
  • Learn how to expand polynomials, specifically using the binomial theorem.
  • Explore the formal definition of the derivative and its applications in calculus.
  • Practice simplifying difference quotients for various polynomial functions.
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Students studying calculus, educators teaching differential calculus, and anyone seeking to understand the foundational concepts of derivatives and limits.

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Simplify the differential qoutient like this (f(x+h)-f(x)) / h
and the qoute is f(x) = 5x^2

Please Help I don't know how to do this!
 
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Didn't you forget a limit for h going to zero? Put in the function then...
 
Izekid said:
Simplify the differential qoutient like this (f(x+h)-f(x)) / h
and the qoute is f(x) = 5x^2
Please Help I don't know how to do this!

You mean "difference quotient". You don't yet have a "differential" because, as TD said, you haven't taken the limit.

All we can say is "do what you are told!"

f(x)= 5x^2 so f(x+h)= 5(x+h)^2. Calculate that.
f(x+h)- f(x)= what you just did, minus 5x^2.

Last thing to do is divide by h.
 

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