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Solve by first point y=x3

Mathematics

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Answer

3x23x^2

Graph of y = x³

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Key points of y = x³
Pointxy
x-intercept00

Step-by-step solution

4 steps

Pour résoudre y=x3y = x^3 par les premiers principes, nous utilisons la définition de la dérivée :

dydx=lim⁡h→0f(x+h)−f(x)h\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Étant donné f(x)=x3f(x) = x^3.

  1. Calculer f(x+h)f(x+h).

    f(x+h)=(x+h)3f(x+h) = (x+h)^3

    Nous développons (x+h)3(x+h)^3 en utilisant la formule du binôme (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3:

    f(x+h)=x3+3x2h+3xh2+h3f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3
  2. Calculer f(x+h)−f(x)f(x+h) - f(x).

    f(x+h)−f(x)=(x3+3x2h+3xh2+h3)−x3f(x+h) - f(x) = (x^3 + 3x^2h + 3xh^2 + h^3) - x^3 f(x+h)−f(x)=3x2h+3xh2+h3f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3
  3. Diviser par hh.

    f(x+h)−f(x)h=3x2h+3xh2+h3h\frac{f(x+h) - f(x)}{h} = \frac{3x^2h + 3xh^2 + h^3}{h}

    Nous factorisons hh au numérateur :

    f(x+h)−f(x)h=h(3x2+3xh+h2)h\frac{f(x+h) - f(x)}{h} = \frac{h(3x^2 + 3xh + h^2)}{h} f(x+h)−f(x)h=3x2+3xh+h2\frac{f(x+h) - f(x)}{h} = 3x^2 + 3xh + h^2
  4. Appliquer la limite quand h→0h \to 0.

    dydx=lim⁡h→0(3x2+3xh+h2)\frac{dy}{dx} = \lim_{h \to 0} (3x^2 + 3xh + h^2)

    Lorsque hh tend vers 0, les termes contenant hh deviennent nuls :

    dydx=3x2+3x(0)+(0)2\frac{dy}{dx} = 3x^2 + 3x(0) + (0)^2 dydx=3x2\frac{dy}{dx} = 3x^2

    La dérivée de y=x3y = x^3 par les premiers principes est 3x2\boxed{3x^2}.

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