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 implicit functions_compressed
Differentiation of implicit functions
Now the first question is: what is an implicit function?
Any function which looks like but not the more common is an implicit function.
For example, is an implicit function.
But is not an implicit function. This is called an explicit function.
In any implicit function, it is not possible to separate the dependent variable from the independent one.
Now I will solve an example of the differentiation of 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 example.
According to Stroud and Booth (2013)* “Find when ”
Here the given function is
Now to get the value of , I’ll differentiate the equation (1).
Here again, I’ll use the standard formulas in differentiation.
Differentiating equation (1) throughout with respect to , I get
Now here both and are single functions.
So it’s quite straightforward to differentiate them.
But the third term is a product of two functions.
For that, I’ll use the product rule of differentiation.
Thus it will be
My next step will be to rearrange the terms.
Now I’ll rearrange the terms to get
Here I can see that ‘3’ is common to each term.
Next, I’ll cancel ‘3’ from each term to get
Now I’ll separate terms with
Thus it will be like
At the end, I’ll simply rearrange the terms to get the final value of as
Thus the value of is
This is the answer to the given example. Now I’ll give another example.
According to Stroud and Booth (2013)* “If , prove that .”
And I have to prove the relation . So, this means I have to differentiate twice. Thus I’ll start with the differentiation.
Therefore, I’ll differentiate both sides of equation (2) with respect to to get
Now that gives
Then I’ll differentiate both sides of equation (3) with respect to . Again I’ll use the product rule of differentiation just like the Example 1. So that gives
Now I’ll prove the relation .
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
Now from equation (3), I can say that
So equation (5) becomes
Next, 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!!