Ross Barker
05/31/2024 · Junior High School
\( \begin{array}{ll}\text { d) } y=4 \text { su derivada es } & \frac{d(4)}{d x}= \\ \text { e) } y=\sqrt{8} \text { su derivada es } & \frac{d(\sqrt{8})}{d x}= \\ \text { f) } y=-\sqrt{15} \text { su derivada es } & \frac{d(-\sqrt{15})}{d x}= \\ \text { g) } y=-\pi \text { suderivada es } & \frac{d(-\pi)}{d x}=\end{array} \)
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Step-by-step Solution
Para encontrar la derivada de las funciones dadas, primero necesitamos recordar las reglas básicas de derivación.
Para la función \( y = 4 \), la derivada es \( \frac{d(4)}{dx} = 0 \) porque la derivada de una constante es cero.
Para la función \( y = \sqrt{8} \), la derivada es \( \frac{d(\sqrt{8})}{dx} = \frac{1}{2\sqrt{8}} \) porque la derivada de \( \sqrt{8} \) es \( \frac{1}{2\sqrt{8}} \).
Para la función \( y = -\sqrt{15} \), la derivada es \( \frac{d(-\sqrt{15})}{dx} = -\frac{1}{2\sqrt{15}} \) porque la derivada de \( -\sqrt{15} \) es \( -\frac{1}{2\sqrt{15}} \).
Para la función \( y = -\pi \), la derivada es \( \frac{d(-\pi)}{dx} = 0 \) porque la derivada de una constante es cero.
Por lo tanto, las derivadas de las funciones dadas son:
- \( \frac{d(4)}{dx} = 0 \)
- \( \frac{d(\sqrt{8})}{dx} = \frac{1}{2\sqrt{8}} \)
- \( \frac{d(-\sqrt{15})}{dx} = -\frac{1}{2\sqrt{15}} \)
- \( \frac{d(-\pi)}{dx} = 0 \)
Quick Answer
- \( \frac{d(4)}{dx} = 0 \)
- \( \frac{d(\sqrt{8})}{dx} = \frac{1}{2\sqrt{8}} \)
- \( \frac{d(-\sqrt{15})}{dx} = -\frac{1}{2\sqrt{15}} \)
- \( \frac{d(-\pi)}{dx} = 0 \)
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