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Let $F_1(x_1,x_2,x_3)=f(x_1,f(x_1,x_2,x_3),x_3)$ and $F_2(x_1,x_2,x_3)=f(x_1,x_2, f(x_1,x_2,x_3))$. Find $\displaystyle \frac{\partial F_i}{\partial x_j}$ for all $i=1,2$ and $j=1,2,3$.

I know that here I have to use chain rule. But since they are not explicit expressions it is very confusing me to find them. Can somebody please explain me how to find them?

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  • $\begingroup$ Try to expand differential $dF_1$ $\endgroup$ Jul 26, 2015 at 16:32

2 Answers 2

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Let's just look at the first case $F_1$. For the sake of making the notation a bit easier, I would introduce new 'coordinates': $$ y_1 = x_1,~~ y_2 = f(x_1, x_2, x_3),~~ y_3 = x_3 $$ We have $F_1(x_1, x_2, x_3) = f(y_1, y_2, y_3)$, where the $y$'s are given by the above.

Now it should be easy enough to apply the chain rule: $\partial F_1/ \partial x_j = \sum_i (\partial f/\partial y_i) (\partial y_i/\partial x_j)$.

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To compute $\frac{\partial F_1}{\partial x_j}$ you have to derive each coordinate of $f$ and multiply by the derivative w.r.t. to $x_j$ of "what is inside" that coordinate, then you add all. Let us call $y_1,y_2,y_3$ the variables of $f$ (i.e. $f(y_1,y_2,y_3))$, then

$$\frac{\partial F_1}{\partial x_1}(x_1,x_2,x_3)=\frac{\partial f}{\partial y_1}(x_1,f(x_1,x_2,x_3),x_3)\\+\frac{\partial f}{\partial y_2}(x_1,f(x_1,x_2,x_3),x_3)\cdot \frac{\partial f}{\partial y_1}(x_1,x_2,x_3) $$

$$\frac{\partial F_1}{\partial x_2}(x_1,x_2,x_3)=\frac{\partial f}{\partial y_2}(x_1,f(x_1,x_2,x_3),x_3)\cdot \frac{\partial f}{\partial y_2}(x_1,x_2,x_3) $$

$$\frac{\partial F_1}{\partial x_3}(x_1,x_2,x_3)=\frac{\partial f}{\partial y_2}(x_1,f(x_1,x_2,x_3),x_3)\cdot \frac{\partial f}{\partial y_3}(x_1,x_2,x_3)+\frac{\partial f}{\partial y_3}(x_1,f(x_1,x_2,x_3),x_3) $$

$$\frac{\partial F_2}{\partial x_1}(x_1,x_2,x_3)=\frac{\partial f}{\partial y_1}(x_1,x_2,f(x_1,x_2,x_3))+\\ \frac{\partial f}{\partial y_3}(x_1,x_2,f(x_1,x_2,x_3))\cdot \frac{\partial f}{\partial y_1}(x_1,x_2,x_3) $$

$$\frac{\partial F_2}{\partial x_2}(x_1,x_2,x_3)=\frac{\partial f}{\partial y_2}(x_1,x_2,f(x_1,x_2,x_3))+\\ \frac{\partial f}{\partial y_3}(x_1,x_2,f(x_1,x_2,x_3))\cdot \frac{\partial f}{\partial y_2}(x_1,x_2,x_3) $$

$$\frac{\partial F_2}{\partial x_3}(x_1,x_2,x_3)=\frac{\partial f}{\partial y_3}(x_1,x_2,f(x_1,x_2,x_3))\cdot \frac{\partial f}{\partial y_3}(x_1,x_2,x_3) $$

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