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The composition of differentiable functions is differentiable. In particular, if $x(t)$ and $y(t)$ are differentiable functions of a parameter $t$, and if $f(x,y)$ is a differentiable function of $x$ and $y$, then $f(x(t),y(t))$ is a differentiable function of $t$. We compute its derivative with the chain rule for paths
The reason behind the chain rule is simple. Since $f(x,y)$ is differentiable, we can approximate changes in $f$ by its linearization, so $$\Delta f \approx f_x \Delta x + f_y \Delta y.$$ Dividing by $\Delta t$ and taking a limit as $\Delta t \to 0$ gives the chain rule. For functions of three of more variables, we just add a term for each variable. If $f(x,y,z)$ is a function of three variables, then $$\frac{d f({\bf r}(t))}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}.$$
Implicit Differentiation Revisited. Back in singlevariable calculus, we used the singlevariable chain rule to compute $dy/dx$ for the level set of a function $f(x,y).$ Let's see what that means in terms of partial derivatives. Let $C$ be a curve defined by an equation $f(x,y)=c$, and let ${\bf r}(t)$ be a parametrization of that curve. Since $f({\bf r}(t))=c$ for all $t$, $df/dt=0$. But by the chain rule, $$f_x \frac{dx}{dt} + f_y \frac{dy}{dt} = \frac{df}{dt} = 0,$$ so $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt} = \frac{f_x}{f_y}.$$ Note that the vector $\langle f_y, f_x \rangle$ is tangent to the curve, while the vector $\langle f_x, f_y \rangle$ is perpendicular to $\langle f_y, f_x \rangle$, and so is perpendicular to the curve. The line through $(a,b)$ and perpendicular to the curve is called the normal line.
