Parametric derivative

Definition

Algebraic definition

The parametric derivative $dy/dx$ for a parametric curve $x=f(t),y=g(t)$ at a point $t=t_{0}$ is given as follows, where $f'(t_{0})$ and $g'(t_{0})$ both exist and $f'(t_{0})\neq 0$:

$\!{\frac {dy}{dx}}|_{t=t_{0}}={\frac {g'(t_{0})}{f'(t_{0})}}$

As a general function of $t$, the parametric derivative $dy/dx$ is defined as $g'(t)/f'(t)$.

NOTE: When calculating the general expression for the parametric derivative, before canceling any factors between $g'(t)$ and $f'(t)$, it is important to separate out the cases where that common value is zero. For any points where $f'(t)=0$, the parametric derivative is not defined (the ordinary derivative may still be defined, but we would need another method to calculate it).

MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a $\{\}_{0}$ subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.

Relation with ordinary derivative

Parametric expressions aren't necessarily functions

The notation $dy/dx$ should be used only in the context that $y$ is a function of $x$, i.e., a single value of $x$ gives rise to a single value of $y$. Generally speaking, this is not guaranteed with parametric curves.

For instance, for a parametric curve $x=f(t),y=g(t)$, $y$ is expressible as a function of $x$ only if $g(t)$ is completely determined by $f(t)$. This is the case, for instance, when $f$ is a one-one function. But it's not always the case. Consider a circle:

$x=\cos t,y=\sin t$

In this case, a single value of $x$, in most cases, corresponds to two values of $y$ that are negatives of each other. That's because if $x_{0}=\cos t_{0}$, we have also $x=\cos(-t_{0})$, so both $\sin t_{0}$ and $-\sin t_{0}$ work. The only case where $y$ is unique in terms of $x$ is when $x=1$ and $x=-1$. Geometrically, vertical secant lines intersect the upper semicircle and lower semicircle, and these are the two $y$-values for a given $x$-value.

"Nice" parametric expressions define functions locally at most points

As described above, for a parametric curve, $y$ need not globally be a function of $x$. However, even in the presented example of a circle, for most $t_{0}$, if we restrict $t$ to a small enough open interval around $t_{0}$, $y$ is a function of $x$ for the part of the curve where $t$ is restricted to that interval. So it's "locally" a function. Differentiation being a local operation, it still makes sense to take the derivative at the point. The parametric derivative should therefore be understood as the derivative for the function obtained by taking the local restriction.

In the circle example, the only points at which $y$ is locally not a function of $x$ are the ones at the far left and far right: $x=1$ and $x=-1$ (ironically, these are the only points with unique global $y$-values).

If $f'(t_{0})\neq 0$, the parametric expression defines a function locally around $t_{0}$

If $f'(t_{0})$ is nonzero, then (depending on the sign of $f'$), for a small enough open interval around $t_{0}$, $f$ is an increasing or decreasing function. In particular, within that open interval, $f$ is a one-one function, so no value of $x$ repeats. Thus locally $y$ is a function of $x$ on that interval.

The parametric derivative makes sense when it is defined

We have now come full circle. In the case that $f'(t_{0})\neq 0$, $y$ is locally a function of $x$ around $t_{0}$. In such cases, if $g'(t_{0})$ also exists, the expression ${\frac {g'(t_{0})}{f'(t_{0})}}$ gives the parametric derivative.

The proof that the parametric derivative expression is correct follows from the chain rule for differentiation. As established above, $y$ is locally a function of $x$ around $t_{0}$. Let's call this local function $h$. Locally around $t_{0}$, we have:

$g=h\circ f$

By the chain rule for differentiation:

$g'(t)=h'(f(t))f'(t)$

At $t=t_{0}$, we get:

$g'(t_{0})=h'(f(t_{0}))f'(t_{0})$

Rearranging, we get:

$h'(f(t_{0}))={\frac {g'(t_{0})}{f'(t_{0})}}$

The left side is the expression we are trying to calculate, and the right side is the expression that we want to prove it to be.