# Differentiation of Parametric Functions

Here you will learn differentiation of parametric functions with example.

Let’s begin –

## Differentiation of Parametric Functions

Sometimes x and y are given as functions of a single variable e.g. x = $$\phi$$(t), y = $$\psi$$(t) are two functions of a single variable. In such a case x and y are called parametric functions or parametric equations and it is called the parameter. To find $$dy\over dx$$ in case of parametric functions, we first obtain the relationship between x and y  by eliminating the parameter t and then we differentiate it with respect to x. But, it is not always convenient to eliminate the parameter. Therefore, $$dy\over dx$$ can also be obtained by the formula

$$dy\over dx$$ = $$dy/dt\over dx/dt$$

Example : find $$dy\over dx$$ where x = a{cos t + $${1\over 2} log tan^2 {t\over 2}$$} and y = a sin t

Solution : We have,

x = a{cos t + $${1\over 2} log tan^2 {t\over 2}$$} and y = a sin t

$$\implies$$ x = a{cos t + $${1\over 2} \times 2 log tan{t\over 2}$$} and y = a sin t

$$\implies$$ x = a{cos t + {$$log tan{t\over 2}$$} and y = a sin t

Differentiating with respect to t, we get

$$dx\over dt$$ = a{-sin t + $${1\over tan t/2}(sec^2{t\over 2})\times {1\over 2}$$} and $$dy\over dt$$ = a cost

$$dx\over dt$$ = a{-sin t + $$1\over 2 sin (t/2) cos (t/2)$$} and $$dy\over dt$$ = a cost

$$\implies$$  $$dx\over dt$$ = a{-sin t + $$1\over sint$$} and $$dy\over dt$$ = a cost

$$\implies$$  $$dx\over dt$$ = a{$$-sin^2t + 1\over sint$$} and $$dy\over dt$$ = a cost

$$dx\over dt$$ = a{$$cos^2t\over sint$$} and $$dy\over dt$$ = a cost

$$\therefore$$   $$dy\over dx$$ = $$dy/dt\over dx/dt$$ = $$acost\over {acos^2t\over sint}$$

$$dy\over dx$$ = tan t