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Zorluk: KolayDifferentiation from First Principles

Using differentiation from first principles, what is the derivative of the function f(x)=x25xf(x) = x^2 - 5x with respect to xx?

  1. 2x52x - 5Cevap
  2. B
    2x+52x + 5
  3. C
    x5x - 5
  4. D
    2x2x

Cevap

The derivative of f(x)=x25xf(x) = x^2 - 5x with respect to xx is 2x52x - 5.
Evaluating the difference quotient f(x+h)f(x)h\frac{f(x+h)-f(x)}{h} for f(x)=x25xf(x) = x^2 - 5x gives x2+2xh+h25x5h(x25x)h=2x+h5\frac{x^2+2xh+h^2-5x-5h-(x^2-5x)}{h} = 2x + h - 5. Taking the limit as h0h \to 0 yields 2x52x - 5.

Adım Adım Çözüm

1
Write down the definition of the derivative from first principles.
f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
This is the standard formula for differentiation from first principles.
2
Substitute f(x)=x25xf(x) = x^2 - 5x and f(x+h)=(x+h)25(x+h)f(x+h) = (x+h)^2 - 5(x+h) into the definition.
\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 5(x+h)] - (x^2 - 5x)}{h}
This sets up the difference quotient for the given polynomial function.
3
Expand the numerator and combine like terms.
\frac{x^2 + 2xh + h^2 - 5x - 5h - x^2 + 5x}{h} = \frac{2xh + h^2 - 5h}{h} = 2x + h - 5
Expanding allows x2x^2 and 5x-5x terms to cancel out, leaving terms containing hh.
4
Evaluate the limit as h0h \to 0.
\lim_{h \to 0} (2x + h - 5) = 2x - 5
As hh approaches 0, the term hh vanishes, giving the final derivative.

Anahtar Kavram

Differentiation from First Principles
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