Calcular derivada derivada de una funciónEjercicios_Resueltosfunciones Calcula la derivada de las siguientes funciones: – a) – b) SOLUCIÓN Comentar el ejercicio
![y = \sqrt[3]{x^5}](local/cache-TeX/ca56e18645e88db42b91ec3df39782eb.png "y = \sqrt[3]{x^5}")
![y = \sqrt[4]{(3x^2-x)^2}](local/cache-TeX/adc6ccda743d2c6a1fff0f387c2c0ad2.png "y = \sqrt[4]{(3x^2-x)^2}")
![\displaystyle \begin{array}{rcl@{\quad}l}f(x) & = & \left(x^{5}\right)^{\dfrac{1}{3}} & \\[4pt] f^{\prime}(x) & = & \left(\left(x^{5}\right)^{\dfrac{1}{3}}\right)^{\prime} & \text{Definición} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{3} \cdot \left(x^{5}\right)^{\dfrac{1}{3} - 1} \cdot \left(x^{5}\right)^{\prime} & \text{Derivada de una potencia (regla de la cadena)} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{3} \cdot \left(x^{5}\right)^{\dfrac{1}{3} - 1} \cdot 5x^{4} & u^{\prime} = 5x^{4} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{3} \cdot \left(x^{5}\right)^{-\dfrac{2}{3}} \cdot 5x^{4} & \text{Simplificamos}\end{array}](local/cache-TeX/8360a6a6a259a1d0f1bd533ce44d6e33.png "\displaystyle \begin{array}{rcl@{\quad}l}f(x) & = & \left(x^{5}\right)^{\dfrac{1}{3}} & \\[4pt] f^{\prime}(x) & = & \left(\left(x^{5}\right)^{\dfrac{1}{3}}\right)^{\prime} & \text{Definición} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{3} \cdot \left(x^{5}\right)^{\dfrac{1}{3} - 1} \cdot \left(x^{5}\right)^{\prime} & \text{Derivada de una potencia (regla de la cadena)} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{3} \cdot \left(x^{5}\right)^{\dfrac{1}{3} - 1} \cdot 5x^{4} & u^{\prime} = 5x^{4} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{3} \cdot \left(x^{5}\right)^{-\dfrac{2}{3}} \cdot 5x^{4} & \text{Simplificamos}\end{array}")
![\displaystyle \begin{array}{l}\begin{array}{rcl@{\quad}l}f(x) & = & \left(\left(3x^{2} - x\right)^{2}\right)^{\dfrac{1}{4}} & \\[4pt] f^{\prime}(x) & = & \left(\left(\left(3x^{2} - x\right)^{2}\right)^{\dfrac{1}{4}}\right)^{\prime} & \text{Definición} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{4} \cdot \left(\left(3x^{2} - x\right)^{2}\right)^{\dfrac{1}{4} - 1} \cdot 2 \cdot \left(3x^{2} - x\right) \cdot \left(6x - 1\right) & \text{Derivada de una potencia (regla de la cadena)} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{4} \cdot \left(\left(3x^{2} - x\right)^{2}\right)^{-\dfrac{3}{4}} \cdot 2 \cdot \left(3x^{2} - x\right) \cdot \left(6x - 1\right) & \text{Simplificamos}\end{array} \\[20pt]\text{Cálculo auxiliar de }u^{\prime}: \\[2pt]\quad \begin{array}{rcl@{\quad}l}u & = & \left(3x^{2} - x\right)^{2} & \\[4pt] u^{\prime} & = & \left(\left(3x^{2} - x\right)^{2}\right)^{\prime} & \text{Definición} \\[4pt] u^{\prime} & = & 2\left(3x^{2} - x\right)^{2 - 1} \cdot \left(3x^{2} - x\right)^{\prime} & \text{Derivada de una potencia (regla de la cadena)} \\[4pt] u^{\prime} & = & 2\left(3x^{2} - x\right)^{2 - 1} \cdot \left(6x - 1\right) & u^{\prime} = 6x - 1 \\[4pt] u^{\prime} & = & 2 \cdot \left(3x^{2} - x\right) \cdot \left(6x - 1\right) & \text{Simplificamos}\end{array}\end{array}](local/cache-TeX/818e86392f092ebf21ff8caa7b9a9b6b.png "\displaystyle \begin{array}{l}\begin{array}{rcl@{\quad}l}f(x) & = & \left(\left(3x^{2} - x\right)^{2}\right)^{\dfrac{1}{4}} & \\[4pt] f^{\prime}(x) & = & \left(\left(\left(3x^{2} - x\right)^{2}\right)^{\dfrac{1}{4}}\right)^{\prime} & \text{Definición} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{4} \cdot \left(\left(3x^{2} - x\right)^{2}\right)^{\dfrac{1}{4} - 1} \cdot 2 \cdot \left(3x^{2} - x\right) \cdot \left(6x - 1\right) & \text{Derivada de una potencia (regla de la cadena)} \\[4pt] f^{\prime}(x) & = & \dfrac{1}{4} \cdot \left(\left(3x^{2} - x\right)^{2}\right)^{-\dfrac{3}{4}} \cdot 2 \cdot \left(3x^{2} - x\right) \cdot \left(6x - 1\right) & \text{Simplificamos}\end{array} \\[20pt]\text{Cálculo auxiliar de }u^{\prime}: \\[2pt]\quad \begin{array}{rcl@{\quad}l}u & = & \left(3x^{2} - x\right)^{2} & \\[4pt] u^{\prime} & = & \left(\left(3x^{2} - x\right)^{2}\right)^{\prime} & \text{Definición} \\[4pt] u^{\prime} & = & 2\left(3x^{2} - x\right)^{2 - 1} \cdot \left(3x^{2} - x\right)^{\prime} & \text{Derivada de una potencia (regla de la cadena)} \\[4pt] u^{\prime} & = & 2\left(3x^{2} - x\right)^{2 - 1} \cdot \left(6x - 1\right) & u^{\prime} = 6x - 1 \\[4pt] u^{\prime} & = & 2 \cdot \left(3x^{2} - x\right) \cdot \left(6x - 1\right) & \text{Simplificamos}\end{array}\end{array}")