Consulta 1b

Re: Consulta 1b

de Rafael Parra -
Número de respuestas: 0
Hola Gonzalo, si consideramos  h(x,y)= g(f(x,y)) , el polinomio de Taylor de la función  h:\mathbb{R^2}\rightarrow \mathbb{R} esta dado por 
h(0,0) + \frac{\partial h}{\partial x}(0,0)x + \frac{\partial h}{\partial y}(0,0)y + \frac{1}{2}\frac{\partial^2 h}{\partial x^2}(0,0)x^2 + \frac{\partial^2 h}{\partial x \partial y}(0,0)xy + \frac{1}{2}\frac{\partial^2 h}{\partial y^2}(0,0)y^2.
Donde 
h(0,0)= g(f(0,0))
\frac{\partial h}{\partial x}(0,0)= g'(f(0,0))\frac{\partial f}{\partial x}(0,0)
\frac{\partial h}{\partial y}(0,0)= g'(f(0,0))\frac{\partial f}{\partial y}(0,0)

Para las segundas derivadas parciales, se tiene:

\frac{\partial^{2} h}{\partial x^{2}}(x,y)= \frac{\partial }{\partial x}( \frac{\partial h}{\partial x} ) = \frac{\partial }{\partial x}(g'(f(x,y)) \frac{\partial f}{\partial x}(x,y))
= g''(f(x,y)) \frac{\partial f }{\partial x}(x,y) + g'(x,y) \frac{\partial ^2 f}{\partial x^2}(x,y)

las restantes se tratan de manera similar