# 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 \[ \frac{d g}{dt}(t_0)=\frac{\partial f}{\partial x}(x(t_0),y(t_0))\cdot \frac{dx}{d t}(t_0)+ \frac{\partial f}{\partial y}(x(t_0),y(t_0))\cdot \frac{dy}{d t}(t_0).\]

We will prove the existence of the limit
\[ \lim_{h\to 0}\frac{g(t_0+h)-g(t_0)}{h}=\lim_{h\to 0}\frac{f(x(t_0+h),y(t_0+h))-f(x(t_0),y(t_0))}{h}.\]
Notice that if we prove the existence of both of the limits
\[
\lim_{h\to 0}\frac{f(x(t_0+h),y(t_0+h))-f(x(t_0+h),y(t_0))}h
\;\;\;\mbox{and}\;\;\;\lim_{h\to 0}\frac{f(x(t_0+h),y(t_0))-f(x(t_0),y(t_0))}h,
\]
then the required result follows immediately, \( g^{\prime}(t_0) \) being the sum of the last two limits.

For the second limit we can apply the theory from the single-variable calculus. The limit in question is the derivative of the function \( f(x(t), y(t_0)) \) at \( t=t_0 \) and according to the chain rule it is equal to \( \frac{\partial f}{\partial x}(x(t_0),y(t_0))\cdot x^{\prime}(t_0) \).

Evaluating the first limit is very similar, although more annoying. There are too many \( h \)s, so we won’t do this. For those who want to try, the hint is: use the Taylor’s theorem (with ``little o’’ remainder) from the single-variable calculus. The resulting limit is \( \frac{\partial f}{\partial y}(x(t_0),y(t_0))\cdot y^{\prime}(t_0) \).

## 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:
\[ \frac{\partial f}{\partial x}(x_0,y_0)=-\frac{\frac{\partial F}{\partial x}(x_0,y_0)}{\frac{\partial F}{\partial z}(x_0,y_0)}\quad\quad\quad \mbox{and}\quad\quad\quad
\frac{\partial f}{\partial y}(x_0,y_0)=-\frac{\frac{\partial F}{\partial y}(x_0,y_0)}{\frac{\partial F}{\partial z}(x_0,y_0)}.
\]

We will skip the proof of the existence of the solution \( f \). Assuming that the solution exists, we will prove the equalities for \( f_x \) and \( f_y \). We do this by differentiating the equation \( F(x,y,f(x,y))=F(x_0,y_0,f(x_0,y_0)) \) with respect to \( x \) and \( y \). The right-hand side is constant, so its derivative is \( 0 \). Hence the chain rule gives:
\[ F_x+F_z\cdot f_x=0\;\;\;\Rightarrow\;\;\; f_x=-\frac{F_x}{F_z}.\]
Similarly we obtain the other identity.