With implicit differentiation we try to find the derivatives of what are called implicit functions. Up to this point you probably never heard this term. This is because we haven't dealt with them in the problems we've been considering.
Until now, we’ve been calculating derivatives of functions that are not implicit. This term is not used normally, but we could call them explicit functions. For example, consider the function defined by the equality:
We can say that this is an explicit function because the dependence on the variable x is very clear. To be more concrete, you can directly calculate the value of f(x) by replacing the value of x in the right side of the equation.
Now, to begin to understand what implicit functions are, let's consider the following equation:
This equation defines a function. In fact, it defines more than one function, but we are interested only in one right now.
Let’s say you give some value to variable x, say x = 2. Replacing that value in the equation I get that
We can solve the equation to get the corresponding value of y. So, "y" is a function of "x", in the sense that y depends on x. However, I needed to put in a little bit of work to obtain the value of y. It is not as easy as taking the square, as in the example of the function above.
This is why we call this type of functions implicit. The function is kind of hidden in the equation.
Let's reconsider the same equation
We know there's a function "y(x)" hidden in there. And let's say we want to find the derivative of that function.
So, first of all, to make clear that y is a function of x, let’s change the symbol y for the symbol f (x). That is, let's write: y = f (x).
Using that notation, we can write our original equation like this:
It will be useful to make a clear distinction now. It is about notation. When we write the symbol dy/dx we’re referring to the derivative of y with respect to x. This symbol is a noun. It represents the derivative (a function).
On the other hand, when we write the symbol d/dx we’re using it as a verb. We refer to the operation of taking the derivative. It’s like a commandment: take the derivative of what follows with respect to x. We’ll call this symbol the derivative operator.
Keeping that in mind, let’s return to our equation.
We can think about this equation as the equality between two functions of x. If these two functions are equal, their derivatives must be equal too.
So, we take the derivative of both sides of the equation
In the left side we can "distribute" the derivative:
And we know that the derivative of 20 (a constant) is zero. The first derivative in the left side is calculated using the power rule:
Now, let's take a look at the derivative that's left. It is the derivative of the function
And then we multiply that by the derivative of the "inner function"
The derivative of the inside is an unknown. It is what we were originally looking for, remember? So, we just use the symbol f'(x) to represent it.
Replacing this in the equation:
Now it is just algebra. We can divide both sides by two:
Now we solve for f'(x):
And now we replace back y=f(x)
And that's the answer. Notice that you could have arrived to this solution using explicit differentiation. That is, you could had solved the original equation for y and applied the chain rule.
For this particular equation you could have solved for y. There are many cases where that isn't possible, or practical, though.
There is a subtle detail in implicit differentiation that can be confusing. It is the fact that when you are taking the derivative, there is composite function in there, so you should use the chain rule.For example, in the equation we just condidered above, we assumed y defined a function of x. So, if y is a function of x, what is y2? It is also a function of x, but a composite function.
We made the fact obious by writing it in function notation. With practice you'll be able to skip this step of writing y as a function. For example, applying the chain rule directly to the previous equation, we could write
Let's see more examples.
Let's consider the equation:
We can assume this equation defines "y" as a function of "x". Just give x any value and you'll get the corresponding value of y.
However, there's no easy way of expressing y = f(x) explicitly. We'll use implicit differentiation on this function.
We apply the derivative operator to both sides:
As we're just starting with implicit differentiation, let's write y in function notation again:
This makes it clear taht we're dealing with a composite function in there. As you gain confidence and experience you won't need to do this anymore.
So, the first term we know how to derive:
And "a" is a constant, so the right side is zero:
And we apply the chain rule in the second term
Now it is algebra
And f'(x) equals
We can replace back f(x)=y
To bring home what you've been learning you need to solve some calculus problems on your own. Consider purchasing The Calculus Problems Manual. This is a series of downloadable eBooks that contain the essential calculus problems you need to know how to solve.
Each volume is affordable (7$) and you can have it immediatelly (direct download). They are also printable, so you can easily print them if you prefer to work on paper.
So, if you want to practice with the type of problems teachers and professors like to put on tests, go to: The Calculus Problems Manual.
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