Differentiate parametric functions. Today I will talk about how to differentiate parametric functions. Have a look!!

If you’re looking for more in the differentiation of functions, do check-in:

**Logarithmic differentiation of functions**

**Differentiation of implicit functions**

Differentiation of parametric functions_compressed

**Differentiate parametric functions**

First of all, I’ll explain what is a parametric function?

###### What is a parametric function?

We all know the standard form of any function is

Now sometimes there are functions like and . In this case both the functions and are dependent on the factor

These kind of equations are called parametric equations.

So I can say that the general form of parametric equations of two variables, say and are Here is the independent variable.

In the same way, the general form of parametric equations of three variables, say and are Here also is the independent variable.

Now the next question is, how to differentiate parametric functions.

###### How can I differentiate parametric functions?

Well, with any standard function the differentiation of is

But with the parametric equations, it is not possible to get the value of in a straight forward way.

For that, it works as follows.

Let me assume the parametric equations of two variables as

First of all, I’ll get the values of and .

Then I’ll get the value of as

Now I will solve an example on that.

** Example of differentiation of parametric functions**

Disclaimer: This is NOT my own example. I have chosen it from some book. I have also given the due reference at the end of the post.

So here is the example.

##### Example

According to Stroud and Booth (2013)* “If and , prove that .”

##### Solution

Here the given parametric equations are:

Now to get the value of , I have to get the value of first.

As I have mentioned above, in order to get the value of the first step is to get the get the value of

So I’ll start with that.

**Step 1**

The parametric equation of is .

Here is a logarithmic function. So I can use logarithmic differentiation of functions.

You can also check out: **formulas for differentiation**

Now I’ll differentiate with respect to to get

My next task is to simplify it.

For that I’ll rewrite as

Also I’ll replace with

Thus it will be

This is because

You can also check out **the standard formulas in trigonometry**

Therefore I can say that I have

(1)

Next I’ll find the value of

**Step 2**

The parametric equation of is .

Now I’ll differentiate both sides with respect to to get

Therefore I can say that I have

(2)

Next I’ll get the value of

Thus it will be

(3)

Now I’ll substitute the values of and from equations (1) and (2) to equation (3) to get

Therefore I can say that I have got the value of as

(4)

Next I’ll try to find the value of

**Step 3**

Now I’ll differentiate equation (4) with respect to to get the value of

Thus it will be

Now is a product of two functions and .

Thus I’ll use the product rule of differentiation in this case. Therefore it will be

From equation (1), I alreay know that

Thus will be

Hence the value of will be

Now my last task is to simplify it to get the final value.

**Step 4**

Thus I’ll start the process now.

First of all, I bring at the front.

So it looks like

Here is a common term to both and .

So I’ll take it out now.

Thus will be

Next I’ll replace with

At the end, I’ll reshuffle the terms as

Finally I have proved the relation .

This is the answer to this example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

## Leave a Reply