Higgins Simmons
06/29/2024 · High School
Derivar: \( f(x)=x^{2 x^{4}}+\frac{4^{x}}{x^{2}} \)
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Step-by-step Solution
Para derivar la función \( f(x) = x^{2 x^{4}} + \frac{4^{x}}{x^{2}} \), utilizaremos la regla de la cadena, la regla del producto y la regla del cociente.
1. **Derivada de \( x^{2 x^{4}} \)**:
Usamos la propiedad de la derivada de una función de la forma \( a^{g(x)} \):
\[
\frac{d}{dx} a^{g(x)} = a^{g(x)} \cdot g'(x) \cdot \ln(a)
\]
En este caso, \( a = x \) y \( g(x) = 2x^{4} \). Primero, derivamos \( g(x) \):
\[
g'(x) = 2 \cdot 4x^{3} = 8x^{3}
\]
Entonces, la derivada de \( x^{2 x^{4}} \) es:
\[
\frac{d}{dx}(x^{2 x^{4}}) = x^{2 x^{4}} \cdot 8x^{3} \cdot \ln(x)
\]
2. **Derivada de \( \frac{4^{x}}{x^{2}} \)**:
Usamos la regla del cociente:
\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^{2}}
\]
donde \( u = 4^{x} \) y \( v = x^{2} \).
Primero, derivamos \( u \) y \( v \):
- Para \( u = 4^{x} \):
\[
u' = 4^{x} \ln(4)
\]
- Para \( v = x^{2} \):
\[
v' = 2x
\]
Ahora aplicamos la regla del cociente:
\[
\frac{d}{dx}\left(\frac{4^{x}}{x^{2}}\right) = \frac{(4^{x} \ln(4)) x^{2} - 4^{x} (2x)}{(x^{2})^{2}} = \frac{4^{x} (x^{2} \ln(4) - 2x)}{x^{4}}
\]
3. **Sumamos las derivadas**:
Finalmente, la derivada de \( f(x) \) es:
\[
f'(x) = x^{2 x^{4}} \cdot 8x^{3} \ln(x) + \frac{4^{x} (x^{2} \ln(4) - 2x)}{x^{4}}
\]
Por lo tanto, la derivada de la función \( f(x) \) es:
\[
f'(x) = x^{2 x^{4}} \cdot 8x^{3} \ln(x) + \frac{4^{x} (x^{2} \ln(4) - 2x)}{x^{4}}
\]
Quick Answer
La derivada de \( f(x) = x^{2 x^{4}} + \frac{4^{x}}{x^{2}} \) es:
\[
f'(x) = x^{2 x^{4}} \cdot 8x^{3} \ln(x) + \frac{4^{x} (x^{2} \ln(4) - 2x)}{x^{4}}
\]
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