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Zorluk: ZorDifferentiation from First Principles

Using differentiation from first principles, evaluate the numerical value of the derivative of the polynomial function f(x)=2x3−3x2+4f(x) = 2x^3 - 3x^2 + 4 at x=2x = 2.

Cevap: 12

Cevap

The numerical value of the derivative at x=2x = 2 is 12.
Applying the first principles limit formula lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} to f(x)=2x3−3x2+4f(x) = 2x^3 - 3x^2 + 4 yields f′(x)=6x2−6xf'(x) = 6x^2 - 6x. Evaluating this derivative function at x=2x = 2 yields 6(4)−12=126(4) - 12 = 12.

Adım Adım Çözüm

1
Set up the difference quotient using the definition of differentiation from first principles
f′(x)=lim⁡h→0[2(x+h)3−3(x+h)2+4]−[2x3−3x2+4]hf'(x) = \lim_{h \to 0} \frac{[2(x+h)^3 - 3(x+h)^2 + 4] - [2x^3 - 3x^2 + 4]}{h}
Differentiation from first principles evaluates the limit of the rate of change as the increment hh approaches zero.
2
Expand (x+h)3(x+h)^3 and (x+h)2(x+h)^2 and subtract f(x)f(x)
f(x+h)−f(x)=6x2h+6xh2+2h3−6xh−3h2f(x+h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 6xh - 3h^2
Expanding the terms allows cancellation of all terms not containing hh.
3
Divide by hh and evaluate the limit as h→0h \to 0
f′(x)=lim⁡h→0(6x2+6xh+2h2−6x−3h)=6x2−6xf'(x) = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 6x - 3h) = 6x^2 - 6x
Dividing by hh eliminates the indeterminate form 00\frac{0}{0}, allowing direct substitution of h=0h = 0.
4
Substitute x=2x = 2 into the derivative function f′(x)f'(x)
f′(2)=6(2)2−6(2)=24−12=12f'(2) = 6(2)^2 - 6(2) = 24 - 12 = 12
Evaluating at the given point gives the slope of the tangent line at x=2x = 2.

Anahtar Kavram

Differentiation from first principles using limit of difference quotient
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Konular

Differentiation from First Principles

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