Brewer Stewart
09/12/2023 · High School
Deriva la función \( h(x) = \frac{3x + 1}{2x^2 + 5} \) utilizando la regla del cociente.
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Para derivar la función \( h(x) = \frac{3x + 1}{2x^2 + 5} \) utilizando la regla del cociente, recordamos que si tenemos una función de la forma \( \frac{u(x)}{v(x)} \), su derivada se calcula como:
\[
h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
donde \( u(x) = 3x + 1 \) y \( v(x) = 2x^2 + 5 \).
Primero, derivamos \( u(x) \) y \( v(x) \):
1. Derivada de \( u(x) \):
\[
u'(x) = \frac{d}{dx}(3x + 1) = 3
\]
2. Derivada de \( v(x) \):
\[
v'(x) = \frac{d}{dx}(2x^2 + 5) = 4x
\]
Ahora, aplicamos la regla del cociente:
\[
h'(x) = \frac{(3)(2x^2 + 5) - (3x + 1)(4x)}{(2x^2 + 5)^2}
\]
Simplificamos el numerador:
\[
h'(x) = \frac{6x^2 + 15 - (12x^2 + 4x)}{(2x^2 + 5)^2}
\]
\[
h'(x) = \frac{6x^2 + 15 - 12x^2 - 4x}{(2x^2 + 5)^2}
\]
\[
h'(x) = \frac{-6x^2 - 4x + 15}{(2x^2 + 5)^2}
\]
Por lo tanto, la derivada de la función \( h(x) \) es:
\[
h'(x) = \frac{-6x^2 - 4x + 15}{(2x^2 + 5)^2}
\]
Quick Answer
La derivada de \( h(x) = \frac{3x + 1}{2x^2 + 5} \) es \( h'(x) = \frac{-6x^2 - 4x + 15}{(2x^2 + 5)^2} \).
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