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Zorluk: OrtaDifferentiation from First Principles

Using differentiation from first principles, what is the derivative dydx\frac{dy}{dx} of the function y=3−2x2y = 3 - 2x^2?

  1. −4x-4xCevap
  2. B
    4x4x
  3. C
    −2x-2x
  4. D
    −4x−2-4x - 2

Cevap

−4x-4x
Applying the definition of derivative f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} to f(x)=3−2x2f(x) = 3 - 2x^2 gives −4xh−2h2h=−4x−2h\frac{-4xh - 2h^2}{h} = -4x - 2h. Taking the limit as h→0h \to 0 results in −4x-4x.

Adım Adım Çözüm

1
Express f(x+h)f(x+h) for the function f(x)=3−2x2f(x) = 3 - 2x^2
f(x+h)=3−2(x+h)2=3−2(x2+2xh+h2)=3−2x2−4xh−2h2f(x+h) = 3 - 2(x+h)^2 = 3 - 2(x^2 + 2xh + h^2) = 3 - 2x^2 - 4xh - 2h^2
Substitute (x+h)(x+h) into the place of xx and expand the squared binomial.
2
Set up the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h}
(3−2x2−4xh−2h2)−(3−2x2)h=−4xh−2h2h\frac{(3 - 2x^2 - 4xh - 2h^2) - (3 - 2x^2)}{h} = \frac{-4xh - 2h^2}{h}
Subtract f(x)f(x) from f(x+h)f(x+h) to isolate terms involving increment hh.
3
Divide numerator terms by hh
h(−4x−2h)h=−4x−2h\frac{h(-4x - 2h)}{h} = -4x - 2h
Factor out hh to cancel the denominator.
4
Take the limit as h→0h \to 0
lim⁡h→0(−4x−2h)=−4x\lim_{h \to 0} (-4x - 2h) = -4x
Evaluating the limit gives the exact derivative function dydx\frac{dy}{dx}.

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Differentiation from First Principles
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Differentiation from First Principles

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