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Difficulty: Very hardDifferentiation from First Principles

When evaluating the derivative of the cubic function f(x)=2x3−5xf(x) = 2x^3 - 5x from first principles, which of the following expressions represents the fully simplified difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} prior to taking the limit as h→0h \to 0?

  1. 6x2+6xh+2h2−56x^2 + 6xh + 2h^2 - 5Answer
  2. B
    6x2+6xh+2h2+56x^2 + 6xh + 2h^2 + 5
  3. C
    2x2+2xh+2h2−52x^2 + 2xh + 2h^2 - 5
  4. D
    6x2−56x^2 - 5

Answer

The simplified difference quotient before evaluating the limit is 6x2+6xh+2h2−56x^2 + 6xh + 2h^2 - 5.
Expanding f(x+h)=2(x+h)3−5(x+h)f(x+h) = 2(x+h)^3 - 5(x+h) using the binomial expansion yields 2x3+6x2h+6xh2+2h3−5x−5h2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h. Subtracting f(x)=2x3−5xf(x) = 2x^3 - 5x leaves 6x2h+6xh2+2h3−5h6x^2h + 6xh^2 + 2h^3 - 5h. Dividing each term by hh gives the correct simplified difference quotient 6x2+6xh+2h2−56x^2 + 6xh + 2h^2 - 5.

Step-by-Step Solution

1
Express f(x+h)f(x+h) for the function f(x)=2x3−5xf(x) = 2x^3 - 5x
f(x+h)=2(x+h)3−5(x+h)=2(x3+3x2h+3xh2+h3)−5x−5h=2x3+6x2h+6xh2+2h3−5x−5hf(x+h) = 2(x+h)^3 - 5(x+h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5x - 5h = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h
Substitute (x+h)(x+h) into f(x)f(x) and expand the binomial terms completely.
2
Subtract f(x)f(x) from f(x+h)f(x+h) to find the numerator f(x+h)−f(x)f(x+h) - f(x)
f(x+h)−f(x)=(2x3+6x2h+6xh2+2h3−5x−5h)−(2x3−5x)=6x2h+6xh2+2h3−5hf(x+h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h
Cancel out the common terms 2x32x^3 and −5x-5x.
3
Divide the numerator by hh to form the difference quotient
\frac{6x^2h + 6xh^2 + 2h^3 - 5h}{h} = 6x^2 + 6xh + 2h^2 - 5
Factor out hh from each term in the numerator and divide by hh.

Key Concept

Differentiation from First Principles
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Topics

Differentiation from First Principles

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