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Zorluk: OrtaDifferentiation from First Principles

In differentiating the quadratic function f(x)=2x2+5xf(x) = 2x^2 + 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 evaluating the limit as h→0h \to 0?

  1. 4x+2h+54x + 2h + 5Cevap
  2. B
    4x+h+54x + h + 5
  3. C
    4x−5+2h4x - 5 + 2h
  4. D
    2x+2h+52x + 2h + 5

Cevap

The fully simplified difference quotient is 4x+2h+54x + 2h + 5.
Expanding f(x+h)=2(x+h)2+5(x+h)f(x+h) = 2(x+h)^2 + 5(x+h) gives 2x2+4xh+2h2+5x+5h2x^2 + 4xh + 2h^2 + 5x + 5h. Subtracting f(x)=2x2+5xf(x) = 2x^2 + 5x leaves 4xh+2h2+5h4xh + 2h^2 + 5h. Dividing this expression term-by-term by hh yields 4x+2h+54x + 2h + 5.

Adım Adım Çözüm

1
Evaluate f(x+h)f(x+h) for the function f(x)=2x2+5xf(x) = 2x^2 + 5x.
f(x+h)=2(x+h)2+5(x+h)=2(x2+2xh+h2)+5x+5h=2x2+4xh+2h2+5x+5hf(x+h) = 2(x+h)^2 + 5(x+h) = 2(x^2 + 2xh + h^2) + 5x + 5h = 2x^2 + 4xh + 2h^2 + 5x + 5h
To apply first principles, we must replace every instance of xx with (x+h)(x+h) and expand completely.
2
Form the numerator of the difference quotient, f(x+h)−f(x)f(x+h) - f(x).
f(x+h)−f(x)=(2x2+4xh+2h2+5x+5h)−(2x2+5x)=4xh+2h2+5hf(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 5x + 5h) - (2x^2 + 5x) = 4xh + 2h^2 + 5h
Subtracting f(x)f(x) cancels out the terms 2x22x^2 and 5x5x that do not contain hh.
3
Divide the result by hh to get the difference quotient.
\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 5h}{h} = 4x + 2h + 5
Factoring out hh from the numerator and canceling it with the denominator simplifies the quotient.

Anahtar Kavram

Differentiation from First Principles (Difference Quotient)
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Differentiation from First Principles

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