# 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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