Contents of this Calculus 3 episode:

Explicit functions, Implicit functions, Differentiation of implicit functions, Implicit differentiation rule, Partial derivative, n-variable implicit function.

The function is an explicit function, its derivative, as expected, is .

A function is implicit if y is not expressed, not in the form of y=...

We get an implicit function if we mess up the function, like this:

and then we take the square root, too

So, this is an implicit function.

If now, we have to differentiate this newly created implicit function, we could do that by differentiating both sides of the equation, and treat y as a function.

actually, it is a function, since .

Well, the derivative of the x on the right side is most definitely 1.

The left side is much more exciting. Here we have a composite function:

And then it also has to be multiplied by the derivative of the inside function.

We need , in other words, the derivative of the function that was given in implicit form.

Let’s try to express

Here it is.

Since , if we substitute this to y...

And this is the same as the explicit derivative.

It is fair to ask why we bothered so much with this, if at the end, we got the same result, except it was a lot more complicated.

Well, the answer is that unfortunately there are some functions that have no explicit forms.

This function has an explicit form, so in this case, it was unnecessary to suffer through the implicit differentiation.

But take a look at this one, for instance.

In this case, y cannot be expressed in any way, so we are forced to use implicit differentiation.

So, we differentiate both sides, but let's not forget that y is a function here.

So, for example is a composite function.

Therefore, we differentiate it as a composite:

Take the derivative of the outside function,

multiplied by the derivative of the inside function.

Now let’s see the implicit differentiation.

We differentiate both sides of the equation:

We need the derivative of y, so we collect all terms on one side, and send all others to the other side:

Then we factor out .

and finally, we divide (and conquer!):

Well, this is the derivative of our function that was given in implicit form.

Now let’s see the differentiation rule for implicit functions.

The point of this method is to make our life easier.

It says that if is an implicit function, then its derivative is:

Well, so far, there is nothing encouraging about this... But let’s see how it works in practice.

Here is the implicit function:

where all terms should be collected on one side,

and it should be called F.

Before we fall victims of a fatal mistake, we must make it clear that this is not a two-variable function, but an implicit function.

The difference between and is huge.

Let's see what the difference is.

Function is a two-variable function indeed, and x and y can be given freely, but

is not a two-variable function. , Let's just try to substitute 0 for x and 1 for y.

We will get 2=0, which is not true, so here only one of x and y can vary freely, the other cannot. So that is why this function is a single variable function.

Now, that we clarified all this, let’s see what the formula says.

The formula says that we should differentiate this function and using the customary partial derivation with respect to x and y.

And here is the implicit derivative.

It is exactly the same result as earlier,

only this time it was much simpler.

Now that's what the implicit differentiation rule is good for.

The rule works for more variables, too.

It says that if is a single variable implicit function, then its derivative is:

If is an n-variable implicit function, then the derivative of as an implicit function with respect to variable is:

Let's see an example for this!

This is a two-variable implicit function.

Even though it has three letters: x, y and z, notice that only two of them can be given freely, due to the equation.

In two-variable functions, x and y are usually the variables, so we can treat this function as

Z=something x and y

Let’s differentiate this with respect to x, and with respect to y!

Calculus 3 episode