Added implicit equations.

docs/en/mathematics/multivariable-calculus/differentation.md

@@ -50,6 +50,20 @@ $$
 
 with $\nabla f(\mathbf{a})$ the gradient of $f$.
 
+## Chain rule
+
+*Definition*: let $D \subseteq \mathbb{R}^n$ ($n=2$ for simplicity) and let $f: D \to \mathbb{R}$, also let $g: \mathbb{R} \to \mathbb{R}$ given by
+
+$$
+    g(t) = f\big(\mathbf{x}(t)\big),
+$$
+
+if $f$ is continuously differentiable, then $g$ is differentiable with
+
+$$
+    g'(t) = \big\langle \nabla f\big(\mathbf{x}(t)\big),\; \mathbf{\dot x}(t) \big\rangle.
+$$
+
 ## Gradients
 
 *Definition*: at any point $\mathbf{x} \in D$ where the first partial derivatives of $f$ exist, we define the gradient vector $\nabla$ by

docs/en/mathematics/multivariable-calculus/implicit-equations.md

@@ -0,0 +1,54 @@
+# Implicit equations
+
+*Theorem*: for $D \subseteq \mathbb{R}^2$ (for simplicty), let $f: D \to \mathbb{R}$ be continuously differentiable and $\mathbf{a} \in D$.  Assume
+
+* $f(\mathbf{a}) = 0$, 
+* $\partial_2 f(\mathbf{a}) \neq 0$, nondegeneracy. 
+
+then there exists an $I$ around $a_1$ and an $J$ around $a_2$ such that $\phi: I \to J$ is differentiable and 
+
+$$
+    \forall x \in I, y \in J: f(x,y) = 0 \iff y = \phi(x).
+$$
+
+Now calculating $\phi' (x)$ with the chain rule
+
+$$
+\begin{align*}
+    f\big(x,\phi(x)\big) &= 0, \\
+    \partial_1 f\big(x,\phi(x)\big) + \partial_2 f\big(x,\phi(x)\big) \phi' (x) &= 0,
+\end{align*}
+$$
+
+and we obtain
+
+$$
+    \phi' (x) = - \frac{\partial_1 f\big(x,\phi(x)\big)}{\partial_2 f\big(x,\phi(x)\big)}.
+$$
+
+*Proof*: will be added later.
+
+## General case
+
+*Theorem*: Let $\mathbf{F}: \mathbb{R}^{n+m} \to \mathbb{R}^m$ given by $F(\mathbf{x},\mathbf{y}) = \mathbf{0}$ with $\mathbf{x} \in \mathbb{R}^n$ and $\mathbf{y} \in \mathbb{R}^m$. Suppose $\mathbf{F}$ is continuously differentiable and assume $D_2 \mathbf{F}(\mathbf{x},\mathbf{y}) \in \mathbb{R}^{m \times m}$ is nonsingular. Then there exists in neighbourhoods $I$ of $\mathbf{x}$ and $J$ of $\mathbf{y}$ with $I \subseteq \mathbb{R}^n,\; J \subseteq \mathbb{R}^m$, such that $\mathbf{\phi}: I \to J$ is differentiable and 
+
+$$
+    \forall (\mathbf{x},\mathbf{y}) \in I \times J: \mathbf{F}(\mathbf{x},\mathbf{y}) = \mathbf{0} \iff \mathbf{y} = \mathbf{\phi}(\mathbf{x}).
+$$
+
+Now calculating $D \mathbf{\phi}(\mathbf{x})$ with the generalized chain rule
+
+$$
+\begin{align*}
+    \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) &= \mathbf{0}, \\
+    D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) + D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) D \mathbf{\phi}(\mathbf{x}) &= \mathbf{0}, \\
+\end{align*}
+$$
+
+and we obtain
+
+$$
+    D \mathbf{\phi}(\mathbf{x}) = - \Big(D_2 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big) \Big)^{-1} D_1 \mathbf{F}\big(\mathbf{x},\mathbf{\phi}(\mathbf{x})\big).
+$$
+
+*Proof*: will be added later.