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Let us say $f : \mathbb{R}^{n + 1} \to \mathbb{R}$ is $C_1$ and we have some point where $\partial_yf(x_0,y_o) \neq 0$. By the implicit function theorem, we can write $y = g(x)$ where $g$ is some $C^1$ function from $\mathbb{R^n} \to \mathbb{R}$. Moreover, a common formula comes up, as I have seen but I'm not sure how its derived. It relates the derivative of $g$ with respect to each basis vector, to the derivatives of $f$. Specifically it says that where $y = g(x)$ (i.e where y can be written as a function of $x$) that: $$ \frac{\partial g}{\partial x_i}(x,g(x)) = -\frac{\frac{\partial f}{\partial x_i}(x,g(x))}{\frac{\partial f}{\partial y}(x,g(x))} $$

How is this formula derived? Most books say it is through the chain rule. Moreover, how does this generalize for the implicit function theorem for mappings $\mathbb{R^{m+n}} \to \mathbb{R}^m$? I am slightly confused in this respect.

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For the formula: \begin{align*} f(x,g(x))&=0\\ \dfrac{\partial}{\partial x_{i}}f(\cdot,g(\cdot))&=0\\ \dfrac{\partial}{\partial x_{i}}f(x,g(x))+\dfrac{\partial f}{\partial y}(x,g(x))\cdot\dfrac{\partial g}{\partial x_{i}}(x)&=0, \end{align*} after simplifying, the formula is derived.

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  • $\begingroup$ Sorry, I'm slightly confused on step 3. How does that work exactly? $\endgroup$ – rubikscube09 Jan 29 '18 at 6:03
  • $\begingroup$ That is really the chain rule, you know that the variable of $f$ is $(x_{1},...,x_{n},y)$. $\endgroup$ – user284331 Jan 29 '18 at 6:10
  • $\begingroup$ You can view like this: If $j\ne i$, then $\dfrac{\partial x_{j}}{\partial x_{i}}=0$ and of course $\dfrac{\partial x_{i}}{\partial x_{i}}=1$. And you know that the partial derivative of the composition map $(x_{1},...,x_{n})\rightarrow f(x_{1},...,x_{n},g(x_{1},...,x_{n}))$ is given by $\displaystyle\sum_{j=1}^{n}\dfrac{\partial f}{\partial x_{j}}\dfrac{\partial x_{j}}{\partial x_{i}}+\displaystyle\dfrac{\partial f}{\partial y}\dfrac{\partial g}{\partial x_{i}}$. $\endgroup$ – user284331 Jan 29 '18 at 6:14

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