# Differentiation of Parametric and Implicit Functions

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**DIFFERENTIATION OF PARAMETRIC/IMPLICIT FUNCTIONS**

**(A) ** \(\boxed{{\text{PARAMETRIC}}\,\,{\text{FUNCTIONS}}}\)

Sometimes, when expressing *y* as a function of *x,* one might not use a direction relation between *x *and *y*; instead, one might express both *x* and *y* as functions of a third variable, say *t*:

\[\begin{align}&x = f\left( t \right)\\&y = g\left( t \right) \end{align}\]

In that case, how would \(\frac{{dy}}{{dx}}\) be evaluated?

One option is to eliminate the parameter *t* and obtain a relation involving only *x* and *y*, from which \(\frac{{dy}}{{dx}}\) may be obtained; however, this could lead to cumbersome expressions.

Another alternative can be taken as follows; we rearrange \(\frac{{dy}}{{dx}}\) to involve *t* also:

\[\begin{align}\frac{{dy}}{{dx}} = \,\,\frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\end{align}\]

This relation says that for evaluating the derivative of *y* w.r.t *x*, we evaluate the derivative of *y* and *x* w.r.t the parameter *t*, and then take their ratio.

Let us try this on some examples:

**(i) ** \(x = r\cos \theta {\rm{ }}\;\;\;\;y = r\sin \theta \) ; r is a constant

This parametric relation represents a circle of radius *r*. We will follow both the approaches to determine \(\frac{{dy}}{{dx}}:\)

\( \Rightarrow \) **ELIMINATION: **

Square and add the two relations for *x* and *y* to obtain:

\[\begin{align}&{{x^2} + {y^2} = {r^2}}\\ \Rightarrow \qquad \; y &= \pm \sqrt {{r^2} - {x^2}} \\\Rightarrow \qquad \quad &\frac{{dy}}{{dx}} = \pm \frac{x}{{\sqrt {{r^2} - {x^2}} }}\left(\text{For each } \text{we obtain two }y' \text{values because the curve is a circle} \right)\end{align}\]

\( \Rightarrow \) ** PARAMETRIC DIFFERENTIATION **

\[\begin{align}&\frac{{dx}}{{d\theta }} = - r\sin \theta\;\;\;\; \frac{{dy}}{{d\theta }} = r\cos \theta \\ \Rightarrow \qquad &\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = - \cot \theta \end{align}\]

**(ii) ** \(x = a\cos \theta {\rm{ }}\;\;\;\;y = b\sin \theta \); *a*, *b* are constants

This parametric relation represents an ellipse with major and minor axis 2*a* and 2*b* respectively.

\( \Rightarrow \) **ELIMINATION **

\(\theta \) can easily be eliminated to obtain:

\[\begin{align}&{\rm{ }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\\ \Rightarrow \quad & y = \pm \frac{b}{a}\sqrt {{a^2} - {x^2}} \\ \Rightarrow \quad & \frac{{dy}}{{dx}} = \pm \,\,\frac{b}{a}\,\,\frac{x}{{\sqrt {{a^2} - {x^2}} }}\end{align}\]

\( \Rightarrow \) **PARAMETRIC DIFFERENTIATION **

\[\begin{align}& {\rm{ }}\frac{{dx}}{{d\theta }} = - a\sin \theta {\rm{\;\;\; }} \frac{{dy}}{{d\theta }} = b\cos \theta \\ \Rightarrow \qquad & \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{ - b}}{a}\,\,\cot \theta \end{align}\]

In later examples, we will observe that in many cases, parametric differentiation turns out to be much more convenient than differentiation after elimination.

**(B) ** \(\boxed{{\text{IMPLICIT}}\,\,{\text{FUNCTIONS}}}\)

Sometimes, the relation between the variables *x* and *y* is specified in the form *f*(*x*, *y*) = 0 that is, *y* is not explicitly specified in terms of *x*, since this explicit expression is either not possible or not convenient.

In such a case, *y* is said to be an implicit function of *x*.

How do we find \(\frac{{dy}}{{dx}}\) in such a case?

We simply differentiate the relation *f*(*x*, *y*) = 0 with respect to *x*, using \(\frac{{dy}}{{dx}}\) for the derivative of the variable *y*. Then we solve for \(\frac{{dy}}{{dx}}\) .

This will become clear from some examples:

\[ \Rightarrow {\rm{ }}{x^2} + {y^2} = 1\]

Differentiating both sides w.r.t *x*:

\[\begin{align}&2x + 2y\frac{{dy}}{{dx}} = 0\\ \Rightarrow \qquad & \frac{{dy}}{{dx}} = \frac{{ - x}}{y}\\ \Rightarrow \qquad &{x^3} + {y^3} + 2xy = 2\end{align}\]

Differentiating both sides w.r.t *x*:

\[\begin{align}&3{x^2} + 3{y^2}\frac{{dy}}{{dx}} + 2x\frac{{dy}}{{dx}} + 2y = 0\\

\Rightarrow \qquad &\frac{{dy}}{{dx}} = \frac{{ - \left( {3{x^2} + 2y} \right)}}{{3{y^2} + 2x}}\\ \Rightarrow \qquad &y = \cos \left( {x + y} \right) \end{align}\]

Differentiating both sides w.r.t *x*:

\[\begin{align}&{\rm{ }}\frac{{dy}}{{dx}} = - \sin \left( {x + y} \right)\left( {1 + \frac{{dy}}{{dx}}} \right)\\ \Rightarrow \qquad & \frac{{dy}}{{dx}} = \frac{{ - \sin \left( {x + y} \right)}}{{1 + \sin \left( {x + y} \right)}}\end{align}\]

Observe that in case of differentiation of implicit functions, the expression for the derivative \(\frac{{dy}}{{dx}}\) will generally not be independent of *y*.

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school