Matthews Long
06/25/2024 · Primary School
2. (20 puntos) Si \( f \) es tres veces diferenciable y \( f^{\prime}(x) \neq 0 \), la derivada de Schwarz de \( f \) en \( x \) se define mediante: \[ \mathfrak{D} f(x)=\frac{f^{\prime \prime \prime}(x)}{f^{\prime}(x)}-\frac{3}{2} \cdot\left(\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}\right)^{2} \] Demuestre que: \[ \mathfrak{D}(f \circ g)=[\mathfrak{D} f \circ g] \cdot g^{\prime 2}+\mathfrak{D} g . \] Nota: Suponga \( h=f \circ g \) y obtenga \( \mathfrak{D} h \) empleando la regla de la cadena. Tenga cuidado al derivar de
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Para demostrar la relación, definimos \( h = f \circ g \) y calculamos \( \mathfrak{D} h(x) \) usando la regla de la cadena. Después de derivar \( h'(x) \), \( h''(x) \) y \( h'''(x) \), sustituimos estas derivadas en la expresión para \( \mathfrak{D} h(x) \) y simplificamos. Finalmente, obtenemos que \( \mathfrak{D}(f \circ g) = [\mathfrak{D} f \circ g] \cdot g^{\prime 2} + \mathfrak{D} g \).
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