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Here is the problem

"Assume that the functions
$I :\mathbb{R}^3 \rightarrow \mathbb{R} $
$F,g : \mathbb{R}^2 \rightarrow \mathbb{R}^+$

are differentiable and they satisfies $F(x_1,x_2) = I( x_1, x_2, g(x_1,x_2) )$. Find the partial derivatives of F in terms of those of I and g"

My attempts at the problem ;
Since its the first time i encounter a problem with two variables only while having three component. There is not much idea in my head ; Yet i guess i should start with
$\nabla F = \nabla f$.
Assistant on this problem is very much appreciated ! Thank You!

p.s (Sorry for the poor sign presentation! An amateur indeed!)

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Look at the form of the functions: $I(x_1,x_2,x_3)=w$, $F(x_1,x_2)=z$, $g(x_1,x_2)=k$ with $w,z,k \in \mathbb{R}$ and $x_1,x_2,x_3$ variables. So by definition

$$F(x_1,x_2)=I(x_1,x_2,g(x_1,x_2))$$ but $g(x_1,x_2)=k$ this $k$, off course, depends of $x_1,x_2$. For example, if $g(0,0)=1$ and $I(0,0,1)=2$ then $F(0,0)=I(0,0,g(0,0))=I(0,0,1)=2$.

The key is that $F$ depends of $I$ and $g$

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The function $$F:\quad {\mathbb R}^3\to{\mathbb R}, \qquad (x_1,x_2)\mapsto I\bigl(x_1,x_2, g(x_1,x_2)\bigr)$$ in question is a composition $F=I\circ{\bf g}$ of two maps: an "outer" function $I$ of three variables: $$I:\quad{\mathbb R}^3\to{\mathbb R},\qquad (u_1,u_2,u_3)\mapsto I(u_1,u_2,u_3)\ ,$$ and an "inner" function $${\bf g}:\quad {\mathbb R}^2\to{\mathbb R}^3, \qquad (x_1,x_2)\mapsto (u_1,u_2,u_3):=\bigl(x_1, x_2, g(x_1,x_2)\bigr)\ ,$$ which can be thought of as parametric representation of the graph surface of $g$. According to the multivariate chain rule we then obtain $${\partial F\over\partial x_1}=\sum_{k=1}^3{\partial I\over\partial u_k}\biggr|_{{\bf g}(x_1,x_2)}\cdot{\partial u_k\over\partial x_1}\biggr|_{(x_1,x_2)}= I_{.1}\cdot 1+I_{.2}\cdot0+I_{.3}\cdot g_{.1}\ ,$$ $${\partial F\over\partial x_2}=\sum_{k=1}^3{\partial I\over\partial u_k}\biggr|_{{\bf g}(x_1,x_2)}\cdot{\partial u_k\over\partial x_2}\biggr|_{(x_1,x_2)}= I_{.1}\cdot 0+I_{.2}\cdot1+I_{.3}\cdot g_{.2}\ .$$

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First write $x_3= g(x_1, x_2)$ so that we can write $F(x_1, x_2)= I(x_1, x_2, x_3)$.

Then $\frac{\partial F}{\partial x_1}= \frac{\partial I}{\partial x_1}+ \frac{\partial I}{\partial x_3}\frac{\partial g}{\partial x_1}$

And $\frac{\partial F}{\partial x_2}= \frac{\partial I}{\partial x_2}+ \frac{\partial I}{\partial x_3}\frac{\partial g}{\partial x_2}$

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