There are instances when rather than defining a function explicitly or implicitly we define it using a third variable. This representation when a function y(x) is represented via a third variable which is known as the parameter is a parametric form. A relation between x and y expressible in the form x = f(t) and y = g(t) is a parametric form representation with parameter as t. Now we will concentrate on how to differentiate these functions using parametric differentiation.
As we already know by the chain rule,
\( \frac {dy}{dt} = \frac {dy}{dx} × \frac {dx}{dt} \)
This chain rule can also be rewritten as,
\( \frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}} \)
Also as mentioned earlier y’ = g'(t) = \( \frac {dy}{dt} \space and \space x’ = f'(t) = \frac {dx}{dt} \)
Thus we can say that,
\( \frac{dy}{dx} = \frac{g'(t)}{f'(t)} \)
Rules for solving problems on derivatives of functions expressed in parametric form:
Step i) First of all we write the given functions x and y in terms of the parameter t.
Step ii) Using differentiation find out \( \frac{dy}{dt} \space and \space \frac {dx}{dt} \)
Step iii) Then by using the formula used for solving functions in parametric form i.e.\( \frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}} \)
Step iv) Lastly substituting the values of \( \frac {dy}{dt} \space and \space \frac{dx}{dt} \)
Now let us go through some examples to get a deeper insight for solving functions of parametric form.
Example 1: Find the value of \( \frac {dy}{dx} \)
i) x = sin t , y = \( t^2 \)
ii) x = \( \frac {3}{t^3} , y = 3t^4 + 5 \)
Solution 1:
i) Since this function is represented in parametric function format beforehand therefore we need to find out the value \( \frac {dy}{dt} \space and \frac {dx}{dt} \)
Now, \( \frac {dy}{dt} = 2t \)
\( \frac {dx}{dt} = cost \)
Now with the help of chain rule we can write,
\( \frac {dy}{dx} = \frac {\frac{dy}{dt}}{\frac{dx}{dt}} \)
Now substituting the value of \( \frac {dy}{dt} \space and \space \frac {dx}{dt} \)
\( \Rightarrow \frac {dy}{dx} = \frac {2t}{cost} \)
This is the required solution of the differentiation of the parametric equation.

ii) Now as these functions are already expressed in terms of therefore to find \( \frac {dy}{dx} \)
, we find out \( \frac {dy}{dt}~and~\frac {dx}{dt} \) sepatately. So
\( \frac {dy}{dt} = 12t^3 \)
\( \frac {dx}{dt} = \frac {9}{t^4} \)
We know,
\( \frac {dy}{dx} = \frac {\frac{dy}{dt}}{\frac{dx}{dt}} \)
\( \frac {dy}{dx} = \frac {12t^3}{\frac {9}{t^4}} \)
\( \frac {dy}{dx} = \frac {4t^7}{3} \)
This is the required solution of the differentiation of the parametric equation.
Example 2: Find the value of \( \frac{dy}{dx} \)
Solution 2: The given functions are parametric in nature.
And we know, \( \frac {dy}{dx} = \frac {\frac{dy}{dt}}{\frac{dx}{dt}} \)
\( \frac {dy}{dt} = e^{sin t} × cost \)
\( \frac {dx}{dt} = 9t^2 \)
\( \Rightarrow \frac {dy}{dx} = \frac {e^{sint} × cos t}{9t^2} \)
This is the required solution.
We are now thorough through the concept of parametric function. To learn more derivative of a function visit our appBYJU’s the learning app.
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