Marsh Griffin
01/17/2023 · Elementary School
Determina la derivada de cada función y calcula la derivada en los puntos que se indican. \( \begin{array}{ll}\text { a. } f(x)=-12 x^{7} & (0,0) y(-1,12) \\ \text { b. } g(x)=9 \sqrt[6]{x^{5}} & (1,9) y(4,2 \sqrt[6]{16}) \\ \text { c. } f(x)=\frac{2}{5 \sqrt[4]{x^{5}}} & \left(1, \frac{2}{5}\right) \text { y }(2,2 \sqrt{2}) \\ \text { d. } g(x)=\frac{3 \pi}{2} & \text { en } x=0 y x=2 \\ \text { e. } h(x)=\frac{2}{3} x^{-2} & \text { en } x=-1 y x=1\end{array} \)
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Tutor-Verified Answer
Step-by-step Solution
Vamos a calcular la derivada de cada función y luego evaluaremos la derivada en los puntos indicados.
### a. \( f(x) = -12 x^{7} \)
**Derivada:**
\[
f'(x) = -12 \cdot 7 x^{6} = -84 x^{6}
\]
**Evaluación en los puntos:**
- En \( x = 0 \):
\[
f'(0) = -84 \cdot 0^{6} = 0
\]
- En \( x = -1 \):
\[
f'(-1) = -84 \cdot (-1)^{6} = -84
\]
### b. \( g(x) = 9 \sqrt[6]{x^{5}} \)
**Derivada:**
Primero, reescribimos la función:
\[
g(x) = 9 x^{5/6}
\]
Ahora derivamos:
\[
g'(x) = 9 \cdot \frac{5}{6} x^{-1/6} = \frac{15}{2} x^{-1/6}
\]
**Evaluación en los puntos:**
- En \( x = 1 \):
\[
g'(1) = \frac{15}{2} \cdot 1^{-1/6} = \frac{15}{2}
\]
- En \( x = 4 \):
\[
g'(4) = \frac{15}{2} \cdot 4^{-1/6} = \frac{15}{2} \cdot \frac{1}{\sqrt[6]{4}} = \frac{15}{2} \cdot \frac{1}{2^{1/3}} = \frac{15}{2^{4/3}}
\]
### c. \( f(x) = \frac{2}{5 \sqrt[4]{x^{5}}} \)
**Derivada:**
Primero, reescribimos la función:
\[
f(x) = \frac{2}{5} x^{-5/4}
\]
Ahora derivamos:
\[
f'(x) = \frac{2}{5} \cdot \left(-\frac{5}{4}\right) x^{-9/4} = -\frac{1}{4} x^{-9/4}
\]
**Evaluación en los puntos:**
- En \( x = 1 \):
\[
f'(1) = -\frac{1}{4} \cdot 1^{-9/4} = -\frac{1}{4}
\]
- En \( x = 2 \):
\[
f'(2) = -\frac{1}{4} \cdot 2^{-9/4} = -\frac{1}{4 \cdot 2^{9/4}} = -\frac{1}{4 \cdot 2^{2.25}} = -\frac{1}{4 \cdot 4 \sqrt{2}} = -\frac{1}{16 \sqrt{2}}
\]
### d. \( g(x) = \frac{3 \pi}{2} \)
**Derivada:**
La derivada de una constante es cero:
\[
g'(x) = 0
\]
**Evaluación en los puntos:**
- En \( x = 0 \):
\[
g'(0) = 0
\]
- En \( x = 2 \):
\[
g'(2) = 0
\]
### e. \( h(x) = \frac{2}{3} x^{-2} \)
**Derivada:**
\[
h'(x) = \frac{2}{3} \cdot (-2) x^{-3} = -\frac{4}{3} x^{-3}
\]
**Evaluación en los puntos:**
- En \( x = -1 \):
\[
h'(-1) = -\frac{4}{3} \cdot (-1)^{-3} = -\frac{4}{3} \cdot (-1) = \frac{4}{3}
\]
- En \( x = 1 \):
\[
h'(1) = -\frac{4}{3} \cdot 1^{-3} = -\frac{4}{3}
\]
### Resumen de resultados:
- a. \( f'(0) = 0 \), \( f'(-1) = -84 \)
- b. \( g'(1) = \frac{15}{2} \), \( g'(4) = \frac{15}{2^{4/3}} \)
- c. \( f'(1) = -\frac{1}{4} \), \( f'(2) = -\frac{1}{16 \sqrt{2}} \)
- d. \( g'(0) = 0 \), \( g'(2) = 0 \)
- e. \( h'(-1) = \frac{4}{3} \), \( h'(1) = -\frac{4}{3} \)
Quick Answer
a. \( f'(0) = 0 \), \( f'(-1) = -84 \)
b. \( g'(1) = \frac{15}{2} \), \( g'(4) = \frac{15}{2^{4/3}} \)
c. \( f'(1) = -\frac{1}{4} \), \( f'(2) = -\frac{1}{16 \sqrt{2}} \)
d. \( g'(0) = 0 \), \( g'(2) = 0 \)
e. \( h'(-1) = \frac{4}{3} \), \( h'(1) = -\frac{4}{3} \)
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