Differentiation of implicit functions. Hello friends, today I will talk about the differentiation of implicit functions. Have a look.

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

**Differentiation of parametric functions**

**Differentiation of logarithmic functions**

**How to differentiate inverse trigonometric functions**

Want to know more about the different rules in differentiation? Please go to:

**Product rule of differentiation**

**Quotient rule of differentiation**

**READ, DOWNLOAD & PRINT – Differentiation of implicit functions (pdf)**

### Differentiation of implicit functions

Now the first question is: what is an implicit function? So the answer is any function which looks like is an implicit function. But the one with the form is not an implicit function.

Now I just give an example, say, . So this is an implicit function. Again is also an implicit function.

But is not an implicit function. Thus it is called an explicit function. Also, in any implicit function, it is not possible to separate the dependent variable from the independent one.

Now I will show how to differentiate an implicit function.

**Examples of the differentiation of implicit functions**

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)* “Find when ”

**Solution**

Now here the given function is

(1)

As I can see, equation (1) represents an implicit function. First of all, I’ll differentiate equation (1) with respect to .

**Step 1**

So I’ll differentiate equation (1) throughout with respect to . Thus I get

Since the differentiation of any constant term is , I can say that will be .

So it will be

Now here both and are single functions. So it’s quite straightforward to differentiate them by using the standard **formulas in differentiation**. But the third term is a product of two functions. So for that, I’ll use the **product rule of differentiation**.

Thus it will be

So this gives

Therefore my next step will be to rearrange the terms.

**Step 2**

Now I’ll rearrange the terms to get

As I can see, here is common to each term. Next, I’ll cancel from each term to get

Now I’ll separate terms with

Thus it will be

So this gives

At the end, I’ll simply rearrange the terms to get the value of as

Thus the value of is

Hence I can conclude that this is the answer to the given example. Now I’ll give another example.

**Example 2**

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

**Solution**

Now here the given function is

(2)

Also, I have to prove the relation . So, this means I have to differentiate twice. Thus I’ll start with the differentiation.

**Step 1**

As I can see, equation (2) is an implicit function since it’s not possible to separate the – and – components. Therefore, I’ll differentiate both sides of equation (2) with respect to .

So this gives

As I can see, is a common term here. So I’ll cancel it throughout. Now that gives

Next, I’ll separate the term. So it becomes

(3)

Then I’ll differentiate both sides of equation (3) with respect to . Again I’ll use the product rule of differentiation just like Example 1. So that gives

(4)

Now I’ll prove the relation .

**Step 2**

As I can see in equation (4), has the coefficient . And I have to prove the relationship where has the coefficient . Hence I’ll multiply equation (4) throughout with to get

(5)

Now from equation (3), I can say that

So equation (5) becomes

Next, I’ll expand and. So that gives

Now I’ll simplify it to get

And I can also rewrite it as

Hence I have proved the relation and this is the answer to the given 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!!

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