Examkin

Soru

Zorluk: OrtaDifferentiation from First Principles

When differentiating the function f(x)=x2−3xf(x) = x^2 - 3x from first principles, what is the simplified expression for the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} for h≠0h \neq 0 before taking the limit as h→0h \to 0?

Cevap: 2x + h - 3 / 2x - 3 + h / h + 2x - 3 / 2x+h-3 / 2x-3+h

Cevap

2x+h−32x + h - 3
Expanding f(x+h)=(x+h)2−3(x+h)=x2+2xh+h2−3x−3hf(x+h) = (x+h)^2 - 3(x+h) = x^2 + 2xh + h^2 - 3x - 3h and subtracting f(x)=x2−3xf(x) = x^2 - 3x yields 2xh+h2−3h2xh + h^2 - 3h. Factoring and dividing by hh gives the simplified difference quotient 2x+h−32x + h - 3.

Adım Adım Çözüm

1
Set up the difference quotient expression
f(x+h)−f(x)h=[(x+h)2−3(x+h)]−(x2−3x)h\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 3(x+h)] - (x^2 - 3x)}{h}
Substitute (x+h)(x+h) into the function definition to evaluate f(x+h)f(x+h).
2
Expand terms in the numerator
(x2+2xh+h2−3x−3h)−x2+3xh\frac{(x^2 + 2xh + h^2 - 3x - 3h) - x^2 + 3x}{h}
Expand (x+h)2=x2+2xh+h2(x+h)^2 = x^2 + 2xh + h^2 and distribute −3-3 across (x+h)(x+h) as well as the negative sign across f(x)f(x).
3
Combine like terms in the numerator
2xh+h2−3hh\frac{2xh + h^2 - 3h}{h}
The terms x2−x2=0x^2 - x^2 = 0 and −3x+3x=0-3x + 3x = 0 cancel out.
4
Divide numerator terms by hh
2x+h−32x + h - 3
Factor out hh from the numerator to get h(2x+h−3)h=2x+h−3\frac{h(2x + h - 3)}{h} = 2x + h - 3 for h≠0h \neq 0.

Anahtar Kavram

Difference Quotient in First Principles Differentiation
Tahmini Süre:1m 30s
Bu soruyu puanla

Konular

Differentiation from First Principles

Bu konudan daha fazla soru