# Parametric and implicit functions

The **tangent** at the point $$(x_0,y_0) = \big(x(t_0),y(t_0)\big)$$ to the parametric curve $$\big(x(t),y(t)\big)$$ is parallel to the vector $$$\big(x'(t_0),y'(t_0)\big).$$$

If $$x'(t_0)\neq 0$$, the parametric derivative or slope at $$t_0$$ is $$$ \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx}= \frac{y'(t_0)}{x'(t_0)}.$$$ It is the slope of the **tangent to the curve** at the point $$\big(x(t_0),y(t_0)\big)$$. If $$x'(t_0) = 0$$ and $$y'(t_0) \ne 0$$, the tangent is vertical.

For instance, the parametric derivative for the circle at $$t_0 = \pi/4$$ is $$$ \frac{dy}{dx} = \frac{\sin'(t_0)}{\cos'(t_0)} = \frac{\cos(t_0)}{-\sin(t_0)}= - \cot(t_0) = -1.$$$

We consider a function defined implicitly by the relation $$$ F\big(y(x),x\big) = 0$$$ around a point $$x_0$$. This means that the value of the function at $$x_0$$ is $$y_0$$ with $$F(y_0,x_0) = 0$$.

The **derivative of an implicit function** is the slope of the tangent to the curve at $$x_0$$. It is obtained by differentiating the implicit equation directly.

For instance, the slope of the tangent to the circle $$ y^2+x^2=1$$ is $$$\displaystyle y'(x) = -x/y.$$$ We get this by using the chain rule to differentiate the equation as a function of $$x$$: $$$ 2yy' + 2x=0. $$$

Functions of $$x$$ are differentiated in the usual way. $$y$$ differentiates to $$y'$$.

The partial derivative is useful for the differentiation of implicit functions.

The partial derivative in $$y$$ of the two-variable function $$ F(y,x)$$ is the derivative of the one-variable function $$y\mapsto F(y,x)$$. Here $$x$$ is frozen and seen as a parameter. The partial derivative is denoted $$\displaystyle \frac{\partial F}{\partial y}$$ or $$F'_y$$.

We define the partial derivative $$F'_x$$ of $$F$$ with respect to $$x$$ similarly.

The **derivative of an implicit function** $$$ F\big(y(x),x\big) = 0$$$ is the slope of the tangent to the curve at $$x_0$$. It is given by the formula $$$ \frac{dy}{dx}(x_0) = \frac{-F'_x\big(y_0,x_0\big)}{F'_y\big(y_0,x_0\big)},\qquad F(y_0,x_0) = 0. $$$

By the chain rule, we get $$$ 0 = \frac{d}{dx} F\big(y(x),x\big) = y'(x) F'_y\big(y(x),x\big) + F'_x\big(y(x),x\big). $$$

For instance, the slope of the tangent to the circle $$ y^2+x^2=1$$ is $$\displaystyle y'(x) = -x/y.$$ We get this by differentiating the equation or computing the partial derivatives $$$ 2yy' + 2x=0,\qquad F'_x = 2x,\quad F'_y = 2y. $$$