derivative for parametric form

Instead of the usual way  $y=f(x)$  to present plane curves it is in many cases more comfortable to express both coordinates, $x$ and $y$, by means of a suitable auxiliary variable, the parametre. It is true e.g. for the cycloid curve.

Suppose we have the parametric form

$x=x(t),y=y(t).$(1)

For getting now the derivative ${\displaystyle \frac{dy}{dx}}$ in a point $P_0$ of the curve, we chose another point $P$ of the curve. If the values of the parametre $t$ corresponding these points are $t_0$ and $t$, we thus have the points  $(x(t_0),y(t_0))$  and  $(x(t),y(t))$  and the slope of the secant line through the points is the difference quotient

${\displaystyle \frac{y(t)-y(t_0)}{x(t)-x(t_0)}}={\displaystyle \frac{\frac{y(t)-y(t_0)}{t-t_0}}{\frac{x(t)-x(t_0)}{t-t_0}}}.$(2)

Let us assume that the functions (1) are differentiable when  $t=t_0$  and that  $x^{\prime }(t_0)\ne 0$. As we let  $t\to t_0$, the left side of (2) tends to the derivative $\frac{dy}{dx}$ and the side to the quotient $\frac{y^{\prime }(t_0)}{x^{\prime }(t_0)}$. Accordingly we have the result

${\left({\displaystyle \frac{dy}{dx}}\right)}_{t=t_0}={\displaystyle \frac{y^{\prime }(t_0)}{x^{\prime }(t_0)}}.$(3)

Note that the (3)may be written

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$$

Example. For the cycloid

$$x=a(\phi -\sin \phi ),y=a(1-\cos \phi ),$$

we obtain

$$\frac{dy}{dx}=\frac{\frac{d}{d\phi }(1-\cos \phi )}{\frac{d}{d\phi }(\phi -\sin \phi )}=\frac{\sin \phi }{1-\cos \phi }=\cot \frac{\phi }{2}.$$