Barber Logan
11/20/2023 · Primary School
Encuentre la derivaa a paritr de la técnica adecuada. \[ \begin{array}{l}h(s)=-2 \sqrt{5 s^{2}+3} \\ \text { a. } h^{\prime}(x)=\frac{x}{\sqrt{5 x^{2}+3}} \\ \text { b. } h^{\prime}(x)=\frac{-10 x}{\sqrt{5 x+3}} \\ \text { c. } h^{\prime}(x)=\frac{10 x}{\sqrt{5 x^{2}+3}} \\ h^{\prime}(x)=\frac{-10 x}{\sqrt{5 x^{2}+3}}\end{array} \]
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Tutor-Verified Answer
Step-by-step Solution
Para encontrar la derivada de la función \( h(s) = -2 \sqrt{5s^2 + 3} \), utilizaremos la regla de la cadena y la regla del producto.
1. **Identificamos la función exterior e interior**:
- La función exterior es \( -2u \) donde \( u = \sqrt{5s^2 + 3} \).
- La función interior es \( u = \sqrt{5s^2 + 3} \).
2. **Derivamos la función exterior**:
\[
\frac{d}{ds}(-2u) = -2 \frac{du}{ds}
\]
3. **Derivamos la función interior**:
Para \( u = \sqrt{5s^2 + 3} \), podemos reescribirlo como \( (5s^2 + 3)^{1/2} \).
Usamos la regla de la cadena:
\[
\frac{du}{ds} = \frac{1}{2}(5s^2 + 3)^{-1/2} \cdot (10s) = \frac{10s}{2\sqrt{5s^2 + 3}} = \frac{5s}{\sqrt{5s^2 + 3}}
\]
4. **Sustituimos en la derivada de la función exterior**:
\[
h'(s) = -2 \cdot \frac{5s}{\sqrt{5s^2 + 3}} = \frac{-10s}{\sqrt{5s^2 + 3}}
\]
Por lo tanto, la derivada de \( h(s) \) es:
\[
h'(s) = \frac{-10s}{\sqrt{5s^2 + 3}}
\]
La respuesta correcta es:
\[
h'(x) = \frac{-10x}{\sqrt{5x^2 + 3}}
\]
Esto corresponde a la opción:
- **d.** \( h'(x) = \frac{-10x}{\sqrt{5x^2 + 3}} \)
Quick Answer
La derivada de \( h(s) \) es \( h'(x) = \frac{-10x}{\sqrt{5x^2 + 3}} \).
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