Change of variables in first-order partial differentiation. Hello friends, today I’ll talk about the change of variables in first-order partial differentiation. Have a look!!

### Change of variables in first-order partial differentiation

Suppose I have three functions and . And the function is dependent on and . Similarly, the functions and are dependent on and .

So it means

Now if I want to find out the value of , I have to use the change of variable method.

Therefore it will be

Similarly, the value of will be

Now I’ll give an example of the change of variables in first-order partial differentiation.

#### An example of the change of variables in first-order partial differentiation

Note: This example does not belong to me. 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 (2011)* “If where and , show that

(a)

(b) .”

##### Solution

Now here I have three functions and . Also, is dependent on and . But both and are functions of and . So, in order to get the value of , I have to use the formula for the change of variables.

Thus the value of will be

Similarly, the value of will be

Therefore I’ll start with the partial differentiation of and . Since both and are dependent on , I’ll use the same technique as the first-order partial differentiation of functions with two variables.

So here comes my first step.

###### Step 1

Here the given functions and are

Now I’ll differentiate partially with respect to to get

(1)

Then I’ll differentiate partially with respect to to get

(2)

Next, I’ll differentiate partially with respect to to get

(3)

And, I’ll differentiate partially with respect to to get

(4)

Therefore, the value of will be

So I can substitute the values of and from equations (1) and (3) respectively to get the value of .

Thus it will be

Now I will rearrange it to get

(5)

Next, I’ll find out the value of .

###### Step 2

So the value of will be

Now I can substitute the values of and from equations (2) and (4) respectively to get the value of .

Thus it will be

Then I will rearrange it to get

(6)

Ok, now the basics are done. So I’ll start working on part (a) of this example.

###### (a)

Now here I have to prove that .

So I’ll start from the right-hand side (RHS) of this relation.

Therefore I’ll substitute the values of and from equations (5) and (6) respectively to the right-hand side of the relation.

Then it will look like

Next, I’ll simplify it to get

Now I’ll rearrange these terms to get

(7)

As I already know from the given example, and

So I substitute these values in equation (??) to get

Now this is the left-hand side part of the given relation.

So I have proved the relation. And this is the answer to part (a) of this example.

Next, comes part (b).

###### (b)

Now here I have to prove that .

So I’ll start from the right-hand side (RHS) of this relation.

Therefore I’ll substitute the values of and from equations (5) and (6) respectively to the right-hand side of the relation.

Then it will look like

Now I’ll simplify it to get

Next, I’ll simplify it further to get

So this means

Now can never be equal to .

Thus the right-hand side of the relationship will be

Now, this is the left-hand side part of the given relation.

So I have proved the relation. And this is the answer to part (b) of this example.

Dear friends, this is the end of my today’s post on the change of variables in first-order partial differentiation. 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!!

Best Maritime universities says

Great article on CHANGE OF VARIABLES IN FIRST-ORDER PARTIAL DIFFERENTIATION