This video is all about implicit differentiation, which is a method that we use to find the
derivative of implicit functions.
Now you're probably wondering what it even means for a function to be called "implicit".
Well, the word "implicit" is the opposite of "explicit", and the word "explicit" basically
means "clearly defined or expressed".
So explicit functions are the ones you've already seen a lot of in your calculus class.
They look like this: a dependent variable, in most cases we call this "y", expressed
as the function of an independent variable, usually "x".
These types of functions can be differentiated using our usual methods without any trouble.
But if a function is not explicit, that is, if y is not clearly expressed as the function
of x, we call it an implicit function.
Implicit functions look like this: x's and y's are tangled up together in such a way
that they are not easily separated.
In this function, for example, we can try to isolate the variables a couple of different
ways, but it just doesn't work.
This is where implicit differentiation comes into play.
Let's go ahead and use it to differentiate this function.
The first thing we'll do is take the derivative of both sides with respect to x.
Notice that the right side of the equation is a constant, and just like always, the derivative
of a constant will always be zero.
On the left side, we'll need to take the derivative of both components, x^2 and y^2.
The x^2 component is easy; we just use power rule to find that the derivative of x^2 is
2x.
And then we come to the y^2, and it seems pretty weird.
How can we differentiate y^2 with respect to x?
Here's what implicit differentiation tells you to do: first, take the derivative as normal,
so here we'll use the power rule again to get 2y.
Then, we need to multiply this term by dy/dx.
The reason for this extra multiplication may seem a little mysterious right now, but we'll
talk about how this is actually just the chain rule in disguise in just a minute.
Once you have the entire function differentiated with respect to x, we just need to use algebra
to solve for dy/dx, which remember is the derivative of y with respect to x.
This is the question we've been working to solve all along: in this case, we've found
that the derivative of y with respect to x is -x/y.
Okay, so now that we've seen that this method works, let's take a closer look at why it
works.
Why do we multiply some parts of the function by dy/dx?
To start to answer this question, let's look at the implicit function we started with.
It's really important to understand that one of these variables is the function of the
other.
This is the "implication" of the implicit function.
Usually, we use x as the independent variable and y as the dependent variable, so y is actually
a function of x.
We could even rewrite the function as x^2+[f(x)]^2 in order to represent this.
Let's keep in mind this substitution.
The function now looks like this, x^2+[f(x)]^2.
But that function of x^2 now looks like a function inside of a function, or a composite
function, which calls for the chain rule.
So let's go ahead and apply the chain rule here.
We'll take the derivative of the outside function, 2f(x), and multiply it by the derivative of
the inside function, which we can denote as f'(x).
And if we substitute y back in for f(x), and denote the derivative of f(x) as dy/dx, you
can see that we end up with the derivative we got using implicit differentiation.
So, implicit differentiation is just a glorified application of the chain rule!
Let's do a quick review of the steps we took to use implicit differentiation.
First, we differentiate both sides of the equation with respect to the independent variable.
Whenever we get to a term that contains the dependent variable, we recognize it as a function
of the independent variable and make sure to multiply it by the derivative with respect
to x.
And finally, we use algebra to solve for the derivative.
It's as simple as that.