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Nostrana [21]
2 years ago
8

La derivada de f(x)=3x^2-5x+2

Mathematics
2 answers:
ludmilkaskok [199]2 years ago
7 0

Answer:

6x-5

Step-by-step explanation:

b^n=nb^{n-1}\\3x^2-5x+2=2*3x^{2-1}-5*1*x^{1-1}+0\\6x-5

MrRa [10]2 years ago
6 0

Answer:

La derivada de la función es:   \bold{f'(x)=6x-5}

Step-by-step explanation:

Recordemos que la derivada de

  • Una suma es igual a la suma de las derivadas:   [a + b + c]'=[a]'+[b]'+[c]'
  • Una constante es cero:   [k]'=0\;\;,\;\;\;\;\text{si}\;\;k=\text{constante}
  • Un producto es:   [ab]' = [a]'b + a[b]'
  • Una potencia:   [a^n]'=na^{n-1}

La derivada de la función f con respecto a x es definida como

   f'(x)=\frac{df}{dx}\\f'(x)=\frac{d}{dx}(3x^2-5x+2)\\f'(x)=\frac{d}{dx}(3x^2)-\frac{d}{dx}(5x)+\frac{d}{dx}(2)

donde

  • \frac{d}{dx}(3x^2)=\frac{d}{dx}(3)\,x^2+3\,\frac{d}{dx}(x^2)=(0)\,x^2+3\,(2x^{2-1})=0+3\,(2x^1)=6x
  • \frac{d}{dx}(5x)=\frac{d}{dx}(5)\,x\;+\;5\,\frac{d}{dx}(x)=(0)\,x+5\,(1x^{1-1})=0+5\,(x^0)=5(1)=5
  • \frac{d}{dx}(2)=0

Por lo tanto,

  f'(x)=\frac{d}{dx}(3x^2)-\frac{d}{dx}(5x)+\frac{d}{dx}(2) \\f'(x)=6x-5+0 \;\;\;\;\;\Rightarrow\;\;\;\;\; \bold{f'(x)=6x-5}

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