Product rule of differentiation. Hello friends, today I will talk about the product rule of differentiation. Here it goes.

### Product rule of differentiation

Suppose is a product of two functions and . This means

Now my task is to differentiate , that is, to get the value of Since is a product of two functions, I’ll use the product rule of differentiation to get the value of Thus will be

See also: Rules for differentiating functions

There are two other rules in differentiation like qoutent rule & chain rule.

Also, I have already written on the differentiation of different kinds of functions like implicit functions, logarithmic functions and parametric functions.

Now I will give some examples.

#### Examples of the product rule of differentiation

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

##### Example 1

According to Stroud and Booth (2013)* “Differentiate .”

##### Solution

###### Step 1

Here the given function is: .

Now is a product of two functions and .

In order to differentiate with respect to , I’ll use the product rule of differentiation.

Thus it will be

So this means

Now this gives

At the end I’ll simplify it to get the value of as

Hence I can conclude that this is the answer to the given example.

Now I will give another example.

##### Example 2

According to Stroud and Booth (2013)* “Prove that satisfies the equation

##### Solution

Here the given function is

As I can see the function is a product of two functions and .

So I have to prove that the function satisfies the equation

Now in this equation, I have both and components. Thus I have the values of both and .

So I’ll start with .

###### Step 1

First of all, I’ll differentiate with respect to . Since is a product of two functions, I’ll use the product rule of differentiation.

So it will look like

Now this gives

Thus it will be

So this gives

Now I’ll simplify it further to get

Next, I’ll differentiate with respect to to get the value of .

Here is a product of two functions as well. Thus once again I’ll use the product rule of differentiation.

Hence it will be

So this means

Thus it will be

Next, I’ll simplify it to get

Now I’ll simplify it further to get the value of as

(1)

As a next step, I’ll rewrite in terms of and

###### Step 2

I have already seen earlier that

(2)

Also the given value of is

(3)

Now I’ll write from equation (1) in terms of and .

For that I’ll use the values of and from equations (2) and (3) respectively.

Thus will be

Therefore I can say that in terms of and is

(4)

Now I’ll go to the next step.

###### Step 3

Finally, I’ll try to prove the given equation

I’ll start from the left-hand side (LHS) of this equation.

So I’ll substitute the value of from equation (4) on the left-hand side (LHS) of the equation.

Thus it will look like

Now I’ll simplify it further to get

Next, I’ll substitute the values of and from equations (2) and (3) respectively to get

Here RHS is the right-hand side of the equation.

Hence finally I have managed to prove the equation.

This is the answer to this example.

Dear friends, this is the end of my 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!!

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