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 2a and 2b 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.