 # Derivatives Of A Function In Parametric Form

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,

$\frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}}$ (Where $\frac {dx}{dt} \ne 0$ )

Also as mentioned earlier  y’ = g'(t) = $\frac {dy}{dt} \space and \space x’ = f'(t) = \frac {dx}{dt}$

Thus we can say that,

$\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ ( where f'(t) $\ne 0$ )

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 $\frac {dy}{dt} \space and \space \frac{dx}{dt}$  and simplify to obtain the result.

Now let us go through some examples to get a deeper insight for solving functions of parametric form.

Example 1: Find the value of $\frac {dy}{dx}$ for the following functions which are expressed in the parametric form.

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 $\frac {dy}{dt} \space and \space \frac {dx}{dt}$ into the above equation we can find the derivative of  w.r.t x

$\Rightarrow \frac {dy}{dx} = \frac {2t}{cost}$

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  $\frac{dy}{dx}$  for y = $e^{sin t}$ and $x = 3t^3$

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 app-BYJU’s the learning app.