Calcular la derivada de:
$z(x)=\dfrac{3x^2-4x}{5x^2-3}$
Solución
| Se tiene: | ||
| $f(x)=(3x^2-4x)$ | ||
| $g(x)=(5x^2-3)$ | ||
| Derivada de una división | ||
| $z'(x)=\dfrac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$ | ||
| $z'(x)=\dfrac {(3x^2-4x)'\cdot(5x^2-3)-(3x^2-4x)\cdot(5x^2-3)'}{(5x^2-3)^2}$ | ||
| $z'(x)=\dfrac {(6x^1-4)\cdot(5x^2-3)-(3x^2-4x)\cdot(10x^1-0)}{(5x^2-3)^2}$ | ||
| $z'(x)=\dfrac {(6x-4)\cdot(5x^2-3)-(3x^2-4x)\cdot(10x)}{25x^4-30x^2+9}$ | ||
| $z'(x)=\dfrac {30x^3-18x-20x^2+12-30x^3+40x^2}{25x^4-30x^2+9}$ | ||
| $z'(x)=\dfrac{20x^2-18x+12}{25x^4-30x^2+9}$ | ||
| Respuesta: | ||
| $z(x)=\dfrac{3x^2-4x}{5x^2-3}$ | ||
| $z'(x)=\dfrac{20x^2-18x+12}{25x^4-30x^2+9}$ | ||