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Zorluk: OrtaDifferentiation from First Principles

By definition, the derivative of the function f(x)=4xf(x) = \frac{4}{x} at x=2x = 2 is given by the limit of the difference quotient lim⁡h→0f(2+h)−f(2)h\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}. What is the value of this limit?

  1. −1-1Cevap
  2. B
    11
  3. C
    −2-2
  4. D
    −4-4

Cevap

−1-1
Substituting f(2+h)=42+hf(2+h) = \frac{4}{2+h} and f(2)=2f(2) = 2 into the difference quotient gives 42+h−2h=−2hh(2+h)=−22+h\frac{\frac{4}{2+h} - 2}{h} = \frac{-2h}{h(2+h)} = \frac{-2}{2+h}. Evaluating the limit as h→0h \to 0 yields −22=−1\frac{-2}{2} = -1.

Adım Adım Çözüm

1
Calculate f(2)f(2) and f(2+h)f(2+h)
f(2)=42=2f(2) = \frac{4}{2} = 2 and f(2+h)=42+hf(2+h) = \frac{4}{2+h}
These are the two values needed for the difference quotient numerator.
2
Subtract f(2)f(2) from f(2+h)f(2+h) and find a common denominator
f(2+h)−f(2)=42+h−2=4−2(2+h)2+h=4−4−2h2+h=−2h2+hf(2+h) - f(2) = \frac{4}{2+h} - 2 = \frac{4 - 2(2+h)}{2+h} = \frac{4 - 4 - 2h}{2+h} = \frac{-2h}{2+h}
Simplifying the numerator expression algebraically.
3
Divide the numerator by hh to form the difference quotient
−2h2+hh=−22+h\frac{\frac{-2h}{2+h}}{h} = \frac{-2}{2+h}
Canceling the common factor hh in the numerator and denominator.
4
Evaluate the limit as h→0h \to 0
\lim_{h \to 0} \frac{-2}{2+h} = \frac{-2}{2+0} = -1
Direct substitution of h=0h = 0 into the simplified expression gives the instantaneous rate of change.

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Differentiation from First Principles

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