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.
Please assume this is my initial exposure to calculus and derivatives. I am having difficulty making the connection between the application of the chain rule to explicit differentiation and that of implicit differentiation. Everything I've learned so far about differentiation has been based on explicitly defined functions and limits. Entry #6: 1.5 Limit Laws & The Squeeze Theorem Entry #7: 2.1 Differentiation Rules, Tangent Lines & Rates of Change Entry #8: 2.2 Product Rule, Quotient Rule & Trig Rules Entry #9: 2.3 Logarithmic Functions & Log Derivatives Entry #10: 2.4 The Chain Rule Entry #11: 3.1 Implicit Differentiation Entry #12: 3.2 Derivatives of Inverse Functions. Some implicit functions can be rewritten as explicit functions. The function y – x 2 = 0 is an implicit function, but it can be rewritten (using basic algebra) as an explicit function as y = x 2. The function y 4 +7y 2x−y 2 x 4 −9x 5 = 3 is an implicit function which cannot be written explicitly. Homework Statement Consider: x^3+y^3+2xy=4, y=1 when x=1 a.) Find the equation of the tangent line to the curve when x=1. B.) Find y' at x=1. C.) Is the graph of y=f(x) concave up or concave down near x=1? Homework Equations Any derivative rules. The Attempt at a Solution For Part a. Implied multiplication has a higher priority than explicit multiplication to allow users to enter expressions, in the same manner as they would be written. For example, the TI-80, TI-81, TI-82, and TI-85 evaluate 1/2X as 1/ (2.X), while other products may evaluate the same expression as 1/2.X from left to right.
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
6.1 Implicit Vs Explicitap Calculus Calculator
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.
Implicit Differentiation
Example 1
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
This is a composite function. So, we must use the chain rule. First, we take the derivative of the 'outer 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.
Fast Implicit Differentiation
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
This is equivalent to what we did before, but this direct method can be confusing initially. Why did the y' appear? Because we applied the chain rule. Why? Because y2 is a composite function of x.Let's see more examples.
Example 2
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
Exercises
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.
Have a Doubt About This Topic? Have an 'Impossible Problem'?
If you have just a general doubt about a concept, I'll try to help you. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. These will appear on a new page on the site, along with my answer, so everyone can benefit from it.
What Other Visitors Have Asked
Click below to see contributions from other visitors to this page...
Implicit and Explicit Differentiation
Given that: Find y' using implicit differentiation . Also find y' writing y as an explicit function of x. Let's first write y as an …
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.
Please assume this is my initial exposure to calculus and derivatives. I am having difficulty making the connection between the application of the chain rule to explicit differentiation and that of implicit differentiation. Everything I've learned so far about differentiation has been based on explicitly defined functions and limits. Entry #6: 1.5 Limit Laws & The Squeeze Theorem Entry #7: 2.1 Differentiation Rules, Tangent Lines & Rates of Change Entry #8: 2.2 Product Rule, Quotient Rule & Trig Rules Entry #9: 2.3 Logarithmic Functions & Log Derivatives Entry #10: 2.4 The Chain Rule Entry #11: 3.1 Implicit Differentiation Entry #12: 3.2 Derivatives of Inverse Functions. Some implicit functions can be rewritten as explicit functions. The function y – x 2 = 0 is an implicit function, but it can be rewritten (using basic algebra) as an explicit function as y = x 2. The function y 4 +7y 2x−y 2 x 4 −9x 5 = 3 is an implicit function which cannot be written explicitly. Homework Statement Consider: x^3+y^3+2xy=4, y=1 when x=1 a.) Find the equation of the tangent line to the curve when x=1. B.) Find y' at x=1. C.) Is the graph of y=f(x) concave up or concave down near x=1? Homework Equations Any derivative rules. The Attempt at a Solution For Part a. Implied multiplication has a higher priority than explicit multiplication to allow users to enter expressions, in the same manner as they would be written. For example, the TI-80, TI-81, TI-82, and TI-85 evaluate 1/2X as 1/ (2.X), while other products may evaluate the same expression as 1/2.X from left to right.
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
6.1 Implicit Vs Explicitap Calculus Calculator
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.
Implicit Differentiation
Example 1
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
This is a composite function. So, we must use the chain rule. First, we take the derivative of the 'outer 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.
Fast Implicit Differentiation
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
This is equivalent to what we did before, but this direct method can be confusing initially. Why did the y' appear? Because we applied the chain rule. Why? Because y2 is a composite function of x.Let's see more examples.
Example 2
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
Exercises
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.
Have a Doubt About This Topic? Have an 'Impossible Problem'?
If you have just a general doubt about a concept, I'll try to help you. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. These will appear on a new page on the site, along with my answer, so everyone can benefit from it.
What Other Visitors Have Asked
Click below to see contributions from other visitors to this page...
Implicit and Explicit Differentiation
Given that: Find y' using implicit differentiation . Also find y' writing y as an explicit function of x. Let's first write y as an …
A Hader Implicit Differetiation Problem
How do we find the second derivative of an implicit function ? The example posted by Eric is the following: Find the second derivative with respect …
Finding Second Derivative of Implicit Function
This is an example of a more elaborate implicit differentiation problem. This is an interesting problem, since we need to apply the product rule in a way …
New! Comments
Do you have a doubt, or want some help with a problem? Leave a comment in the box below.Integration Mini Video Lecture
This video explains the differences and similarities between indefinitie and definite integration.
Video transcript
[Music]
Related lessons on IntMath
This mini video lecture explains the processes in these two pages:
... and ...
What's the difference between indefinite and definite integrals?
Indefinite integral
With an indefinite integral there are no upper and lower limits on the integral here, and what we'll get is an answer that still has x's in it and will also have a K, plus K, in it.
6.1 Implicit Vs Explicitap Calculus Algebra
A definite integral has upper and lower limits on the integrals, and it's called definite because, at the end of the problem, we have a number - it is a definite answer.
OK. Let's do both of them and see the difference.
The integral x to the five dx is equal to ... `(int x^5 dx =?)`
... Now, for integration, I have to add one to the index.
So, it's going to be x to the 6
...and then I divide by the new number ... x to the sixth over 6 `(int x^5 dx = x^6/6...)`
...and then I must remember the plus K. `(int x^5 dx = x^6/6+K)`
K is just some constant. We might have some information elsewhere in the problem that will help us to find the constant. In this case we don't have any extra information, so it's just plus K.
Definite integral
Now, let's see what it looks like as a definite integral, this time with upper and lower limits, and we'll see what happens. `(int_1^2 x^5 dx = ?)`
Step 1 is to do what we just did. We integrate, and I'm going to have once again x to the six over 6, but this time I do not have plus K - I don't need it, so I don't have it.
And I write it like this,
... and then I have a lower limit of 1
... and an upper limit of 2. `(int_1^2 x^5 dx = [x^6/6]_1^2)`
These are the numbers that come from here in the question - 1 comes from here, 2 comes from here.
So, what I've done is I've integrated, and ... written in the upper and lower limits correctly.
The next step is to actually substitute these upper and lower limits.
So, what I can do is like this ... and like this ... and I'm going to substitute 2 into this position, and 1 into this position.
So, what have I done? `(int_1^2 x^5 dx = [x^6/6]_1^2=2^6/6-1^6/6)`
6.1 Implicit Vs Explicitap Calculus 2nd Edition
I've found the integral, and I've substituted 2 into this position ... and 1 into this position ... and I'm going to subtract the difference ... and I'm just going to use Scientific Notebook now to give me that, as a decimal.
6.1 Implicit Vs Explicitap Calculus Solver
And it turns out to be 10.5.
`int_1^2 x^5 dx = [x^6/6]_1^2` `=2^6/6-1^6/6` `=10.5`
And this is a definite answer.
[Music]
