- Implicit differentiation
- Worked example: Implicit differentiation
- Worked example: Evaluating derivative with implicit differentiation
- Implicit differentiation
- Showing explicit and implicit differentiation give same result
- Implicit differentiation review
Sal gives an example of how implicit differentiation results in the same derivative expression as direct differentiation. Created by Sal Khan.
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- It seems to me like it's more work to use implicit differentiation than it is explicit differention. Is there any good reason to use implicit differentiation?(9 votes)
- Sure, you cannot always isolate y or else doing so is nightmarishly difficult. In these cases implicit differentiation is much easier.
For example, try finding the derivative of this by explicit differentiation:
- I understand the mechanics here, I believe I get that this works but not why. The part I am confused about is we now suddenly seem to be doing math WITH operators. I though the d/dx was just an operator symbol, but now we seem to be treating like a number or variable and can multiply or divide with it?!
Clearly I am misunderstanding, but to me it's like suddenly learning " (+) * (+) = +^2", addition sign times addition sign of course equals addition sign squared".(16 votes)
- There is no spoon!
Yes, d/dx is an operator, but not like +
Other than the fact that + is diadic and d/dx is monadic (which has nothing to do with the discussion at hand, but it is a difference between the two) it is best to think of d/dx as an operator that takes a function as input, and returns a function. In this light it may be easier to see how it is being manipulated. This will be even more clear in Integral calculus when you do u-sub problems. Later you will be seeing the same thing with other operators you will be learning in multivariable calculus.
Do your best to accept it and use it (or do your own research on what d/dx, in all its subtle glory, is). When I was a student I used to say that the stuff I was taught in one semester would be understandable (from a theorem point of view, not a use point of view) in the following semester, by which time, our maturity grows so the we can better accept the truth. (which is to say, sometimes we need more time to understand the math within the proofs of the tools we are using, but that we don't understand at this entry level of our math studies, doesn't mean the concepts don't work; a rigorous proof of why does exist)(2 votes)
- at0:59, is dy/dx the same thing as d/dx [x^-2] because y = x^-2?(7 votes)
- yes that's how you write the notation. You can also write dy/dx as (y)' or f'(x). There are many ways in which you can indicate a derivative of a function.(8 votes)
- So in the end, to solve for y, we still had to use the result from explicit differentiation? So is it impossible to solve for y using only implicit differentiation?(7 votes)
- Sometimes you cant solve explicitly. The original equation however could be, he was just using it as an example to prove the explicit and implicit both provide the same answer. However, when you solve implicitly, you get a y variable in the answer, which would have to be replaced with the y= from the original equation. When you do that and simplify, both answers are the same.(5 votes)
- Is there a way to explain implicit differentiation in a way other than “it is an application of the chain rule”? Can it be explained in terms of limits? I am not feeling comfortable with implicit differentiation. I can apply the rules and get the correct answers, but I don’t have a gut feeling for why it works. It is hard for me to explain. Please bear with me.
It feels like a leap to me to say that the chain rule fully explains implicit differentiation. Everything I’ve learned so far about differentiation has been based on explicitly defined functions and limits. Applying the chain rule to explicit functions makes sense to me, as I am just recognizing composite functions within an original function. But applying the chain rule to a non-function, e.g. an equation of a circle, and to the dependent variable, seems like a giant leap. Take for example, the equation x√y=1. I understand that y is a function of x, but it is the given function that makes it a function of x. Solving the equation for y yields y=1/x^2 . If I substitute 1/x^2 in for y in the original equation, I get 1=1. This is different than the equation y=xsin(2x^2+2x+1) where 2x^2+2x+1 is a composite function.
Perhaps I am not completely understanding the chain rule. What am I missing? Thanks for any help you can provide.(6 votes)
- Thinking of d/dx as an operator that accepts functions and returns their derivative is perfectly valid, and you derivation of d/dx[x]=1 is valid.(3 votes)
- What supports Sal squaring both sides at :30(1 vote)
- That's basic algebra. If a = b, then a^2 = b^2. You're multiplying both sides of the equation by amounts that are known to be equal (a and b in my example, or sqrt(y) and 1/x in the video), so the equality is preserved.(12 votes)
- i am confused to what explicit and implicit mean(2 votes)
- explicit differentiation means that you first have to solve for y and find the answer that way. If you solve it implicitly, you kind of do it backwards, using terms that don't have implicit definitions (in this case the derivative of y is part of the equation, even though we don't even know what it means yet.)(4 votes)
- Im confused at the end of the video when Sal explains how explicit and implicit differentiation are related. Could someone clarify the relationships and similarities of the two?(2 votes)
- Finding the derivative explicitly is a two-step process: (1) find y in terms of x, and (2) differentiate, which gives us dy/dx in terms of x.
Finding the derivative implicitly is also two steps: (1) differentiate, and (2) solve for dy/dx. This method may leave us with dy/dx in terms of both x and y. However, we previously found y in terms of x when we found the derivative explicitly. We have an expression in x that is equal to y, so we can substitute that expression for y anywhere y appears in the derivative we found implicitly. The result will be dy/dx in terms of x, eliminating any y that appears on the right side of the equation. If we did everything right, this derivative will be the same as the one we found explicitly.(7 votes)
- I suppose it depends on the equation...but can you just look at an equation and know that it can't be derived using explicited differentiation? Is there a sort of 'telltale' that lets you know that "okay, this literally can't be solved explicitedly, and must be implicitedly derived?"(3 votes)
- @2:42Sal uses the chain rule. Do you always have to use the chain rule in implicit differentiation?(1 vote)
- Yes, implicit differentiation is a special application of the chain rule. It's how we take the derivative of an expression involving y with respect to x, which otherwise doesn't sound possible (we normally need a function of x in order to differentiate with respect to x).(7 votes)
What I want to show you in this video is that implicit differentiation will give you the same result as, I guess we can say, explicit differentiation when you can differentiate explicitly. So let's say that I have the relationship x times the square root of y is equal to 1. This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1/x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. So we get dy dx is equal to negative 2 x to the negative 2 minus 1-- x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing-- x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, well, actually, we're going to apply both the product rule and the chain rule. The product rule tells us-- so we have the product of two functions of x. You could view it that way. So this, the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, times the derivative with respect to x of the square root of y. Let me make it clear, this bracket. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. This simplifies to 1, so we're just going to be left with the square root of y right over here. So this is going to simplify to a square root of y. And what does this over here simplify to? Well the derivative with respect to x of the square root of y, here we want to apply the chain rule. So let me make it clear. So we have plus this x plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of the square root of something with respect to that something. Well, the derivative of the square root of something with respect to that something, or the derivative of something to the 1/2 with respect to that something, is going to be 1/2 times that something to the negative 1/2 power. Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1/2 x to the negative 1/2. Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand side. On the right-hand side, we just have a 0. And now, once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over, so I have, once again, more room to work with. So let me cut it, actually. And then let me paste it. Let me move it over, right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then, if we subtract the square root of y from both sides-- and I'll try to simplify as I go-- we get this thing, which I can rewrite as x times-- well, it's just going to be x in the numerator divided by 2 times the square root of y. y to the negative 1/2 is just the square root of y in the denominator. And 1/2, I just put the 2 in the denominator there-- times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here, just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx-- or dividing both sides by this is the same thing as multiplying by the reciprocal of this-- is equal to 2 times the square root of y over x-- over my yellow x-- times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to y times-- the square root of y times the square root of y is just y. The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2 x to the negative third power. The key here is to realize that this thing right over here we could solve explicitly in terms of-- we could solve for y. So we could just make this substitution back here to see that these are the exact same thing. So if we make the substitution y is equal to 1 over x squared, you would get dy dx, the derivative of y with respect to x, is equal to our negative 2 times 1 over x squared, and then all of that over x, which is equal to negative 2 over x to the third, which is exactly what we have over here, negative 2 x to the negative third power.