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# Parametric and implicit functions

## Differentiation of parametric 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 $$\frac{dy}{dx} = \frac{\sin'(t_0)}{\cos'(t_0)} = \frac{\cos(t_0)}{-\sin(t_0)}= - \cot(t_0) = -1.$$

Parametric equation of a circle

## Differentiation of implicit functions

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 $y'$.

Derivative of the implicit representation of a circle $y^2+x^2=1$

## Implicit functions and partial derivative

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 $$2yy' + 2x=0,\qquad F'_x = 2x,\quad F'_y = 2y.$$