# Parametric & implicit differentiation

As well as basic differentiation and differentiation using the product, quotient or chain rule there is parametric differentiation and implicit differentiation. In the former, we can apply parametric differentiation when two functions are defined parametrically. In the latter, we can apply implicit differentiation when a function is defined implicitly.

## Parametric Differentiation

Recall that parametric equations may look as follows:

,

where is the parameter. See more on parametric equations. In order to find the derivative , we can first find the Cartesian equation or we can use parametric differentiation. We can perform parametric differentiation by noting that and using the derivatives of the parametric equations:

Hence, we find the derivative of with respect to by dividing the derivatives of the original parametric equations. For example, suppose and for . In this case, and . It follows that (since ) . This can also be seen when we find the Cartesian equation first. The Cartesian equation is and so as before (more on differentiating polynomials). See Examples 1 and 2 for more parametric differentiation.

## Implicit Differentiation

If a Cartesian equation is defined implicitly, we can use implicit differentiation to find an expression for . In these cases, we must treat as an unknown function of . For example, consider the implicit equation . We could find the explicit equation and so (see more on differentiating polynomials). However, we can also differentiate implicitly by differentiating both sides of with respect to as it is. The right hand side becomes 0 but the left hand side is a product (treating as an unknown function of ) and so the product rule should be used:

It follows that which we can rearrange to get . This is the same expression that we obtained finding the Cartesian equation first. In most cases at this level it is much more complicated to find the explicit equation so using implicit differentiation makes more sense.

Consider finding an expression for in a second example of an implicit equation: . Both terms on the left are product whereas the right hand side is a polynomial term. Differentiating both sides with respect to gives . Note that is a chain: is an unknown function of that is then squared. Hence, we use the chain rule and the derivative of is . In order to obtain an expression for the derivative we rearrange to make the subject: . We obtain this by keeping all derivative terms on the left and then factorising. See Examples 3 and 4.

## Examples

Find an expression for given the parametric equations and for . Verify your result from the Cartesian equation.

Solution:

Firstly, we can find an expression for using the quotient rule. Let and . It follows that and and so . We can write this in terms of by recalling from Example 1 on the Parametric Equations page that . Hence, . Secondly, we can find an expression for using the chain rule if we write . Let and . It follows that and and so . Hence, .

Recall from Example 1 on the Parametric Equations page that the Cartesian equation is . We can find an expression for using the quotient rule. Let and . It follows that and and so . This is the same expression we obtained using parametric differentiation. In this case, however, it may have been simpler to find the Cartesian equations first.

Find the equation of the tangent to the curve defined by the parametric equations and at the points where .

Solution:

The equation of a tangent takes the form   and so we must find the and coordinates and the gradient at . The and coordinates are and . To find the gradient, we first find the derivatives of the parametric equations: and (see more on differentiating trigonometric functions). It follows that . At ,  the derivative is . Since we have the coordinates and gradient we can find the -intercept from . It follows that and the equation of the tangent is .

In this example, it is much simpler to use parametric differentiation than find a Cartesian equation.

Find the value of the gradient of the curve given implicitly by at the point .

Solution:

Using a combination of the differentiation rules, we can differentiate both sides with respect to to obtain:

.

It follows that

or

.

Hence,

.

Substituting and we have

.

Hence, the gradient at the point is .

Find the exact equation of the normal to the curve defined implicitly by at the point where .

Solution:

In order to find the equation of the normal we first need the gradient when . There are products on both sides of the equation and there is also a chain on the right. So, we use both the product rule and chain rule to differentiate:

.

It follows that

Hence, in order to find the gradient at the point where , we also require the -coordinate. From the original equation, when it follows that or . Hence, and so . The gradient of the normal is then . We find the -intercept from , hence . Hence, the exact equation of the normal at is .