Partial differential of composite function

Posted September 22nd, 2007 by Structure

Let $h(x_1,x_2,...x_n)=g(f_1(x_1,x_2,...x_n),f_2(x_1,x_2,...x_n),....f_m(x_1,x_2,...x_n))$
Suppose that g and f are differentiable functions. Then h is differentiable and partial derivative can be founded by $dh(x)=dg(f(x)df(x)$ . Each differential is a linear application and is represented by the matrix of partial derivatives.For example $J_h(x)=(\frac{\partial h}{\partial x_1},\frac{\partial h}{\partial x_2},....\frac{\partial h}{\partial x_n})$
Following composition rule for matrix, we have

$\frac{\partial h}{\partial x_i}=\sum_{k=1}^{k=m}\frac{\partial g}{\partial y_k}(f)\frac{\partial f_k}{\partial x_i}$
This sum expresses the product of a line in $J_g (f(x))$ with a column in $J_f(x)$
or more explicitly
$\frac{\partial h}{\partial x_i}(x_1,x_2,...x_n)=$
$\sum_{k=1}^{k=m}\frac{\partial g}{\partial y_k}(f_1(x_1,x_2,...x_n),f_2(x_1,x_2,...x_n),....f_m(x_1,x_2,...x_n))\frac{\partial f_k}{\partial x_i}(x_1,x_2,...x_n)$
If g and f are twice differentiable second order partial differential can be express
$\frac{\partial^2 h}{\partial {x_i}_2\partial {x_i}_1}=\sum_{k_2=1}^{k_2=m}\sum_{k_1=1}^{k_1=m}\frac{\partial ^2 g}{\partial {y_k}_2\partial {y_k}_1}(f)\frac{\partial {f_k}_2}{\partial {x_i}_2}\frac{\partial {f_k}_1}{\partial {x_i}_1}+\sum_{k_1=1}^{k_1=m}\frac{\partial g}{\partial {y_k}_1}(f)\frac{\partial ^2{f_k}_1}{\partial {x_i}_2\partial {x_i}_1}$

Partial differential of composite function

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