Chain Rule (Multivariable Calculus)

Chain rule

Theorem (Chain rule)
 
Assume that \( x,y:\mathbb R\to\mathbb R \) are differentiable at point \( t_0 \). Assume that \( f:\mathbb R\times\mathbb R\to\mathbb R \) is differentaible at point \( (x(t_0),y(t_0)) \) with continuous partial derivatives. Then the function \( g(t)=f(x(t),y(t)) \) is differentiable at the point \( (x(t_0),y(t_0)) \) and satisfies \[ \frac{d g}{dt}(t_0)=\frac{\partial f}{\partial x}(x(t_0),y(t_0))\cdot \frac{dx}{d t}(t_0)+ \frac{\partial f}{\partial y}(x(t_0),y(t_0))\cdot \frac{dy}{d t}(t_0).\]

Implicit function theorem

Theorem (Implicit function theorem)
 
Let \( F:\mathbb R^3\to\mathbb R \) be a continuously differentiable function such that \( \frac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0 \). Then the implicit equation \( F(x,y,z)=F(x_0,y_0,z_0) \) has a solution \( z=f(x,y) \) in a neighborhood of the point \( (x_0,y_0) \) and the partial derivatives of \( f \) satisfy: \[ \frac{\partial f}{\partial x}(x_0,y_0)=-\frac{\frac{\partial F}{\partial x}(x_0,y_0)}{\frac{\partial F}{\partial z}(x_0,y_0)}\quad\quad\quad \mbox{and}\quad\quad\quad \frac{\partial f}{\partial y}(x_0,y_0)=-\frac{\frac{\partial F}{\partial y}(x_0,y_0)}{\frac{\partial F}{\partial z}(x_0,y_0)}. \]


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