Log in Jack 11 years agoPosted 11 years ago. Direct link to Jack's post “Would this be considered ...” Would this be considered multivariable calculus? • (185 votes) Ethan Dlugie 11 years agoPosted 11 years ago. Direct link to Ethan Dlugie's post “No, it is not. In multiva...” No, it is not. In multivariable calculus, you have multiple variables which are not related. In this case, the implicit relationship means that y is still a function of x. (301 votes) bruno.garcia 11 years agoPosted 11 years ago. Direct link to bruno.garcia's post “Sorry, maybe it's a silly...” Sorry, maybe it's a silly question, but how do we know when the variables are implicitly related? I mean, what would be an example of non-related variables? • (134 votes) Mehrdad Ghods 11 years agoPosted 11 years ago. Direct link to Mehrdad Ghods's post “Actually it is an interes...” Actually it is an interesting question. Let's start with an example of non-related variables: So, how can we know when the variables are implicitly related? Whenever you can show that changing one variable x necessarily leads to change in the other variable y to hold your equation valid you can say those two variables are related. In short you can define a general dy/dx or dx/dy This might be helpful if you are really into this: (160 votes) Enrico 11 years agoPosted 11 years ago. Direct link to Enrico's post “Am I the only one puzzled...” Am I the only one puzzled by the very first operation? I mean, applying the d/dx operator to both sides of the equation... is it legit? Does it preserve the equality? What's its meaning? Why does it lead to the result that we want(dy/dx of that relationship)? • (108 votes) Ethan Dlugie 11 years agoPosted 11 years ago. Direct link to Ethan Dlugie's post “I understand your concern...” I understand your concerns. At first, it does seem a little sketchy to take the derivative of both sides of an equation. However, you are already familiar with this. If I established the equality (158 votes) Eric Robinson 10 years agoPosted 10 years ago. Direct link to Eric Robinson's post “From 3:08 to 4:51, Sal tr...” From • (31 votes) Just Keith 10 years agoPosted 10 years ago. Direct link to Just Keith's post “Let me cover a little abo...” Let me cover a little about the chain rule that is sometimes missed. When taking any derivative, we always apply the chain rule, but many times that is trivially true and just ignored. For example, d/dx (x²) actually involves the chain rule: We do the same thing with y², only this time we won't get a trivial chain rule Personally, I think using the y' and x' notation is easier for implicit differentiation. Here is the same computation using that notation: (98 votes) Vivian Leung 10 years agoPosted 10 years ago. Direct link to Vivian Leung's post “I don't quite understand ...” I don't quite understand the difference between d/dx and dy/dx. If we're taking the derivative of y with respect to x in this case, what was it that we were doing before? I'm really confused by these terms now... • (42 votes) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “The expression d/dx can b...” The expression d/dx can be taken to mean "the rate of change with respect to x" of whatever follows it. So we can write, for example, d/dx x² = 2x. As another example, we can write d/dx y, and this would mean "the rate of change with respect to x of y." But it's more convenient to combine the d/dx and the y to write dy/dx, which means the same thing. y = x² (62 votes) Emily Phillips 11 years agoPosted 11 years ago. Direct link to Emily Phillips's post “This might be a dumb ques...” This might be a dumb question, but, it appears that finding the tangent line if a point on the unit circle is like finding either it's tangent or cotangent. Is that the case, is that why they're named the way they are? • (26 votes) Matthew Daly 11 years agoPosted 11 years ago. Direct link to Matthew Daly's post “Yup. Here is a diagram t...” Yup. Here is a diagram that I made that shows the motivation for naming both the tangent and secant functions. The cotangent and cosecant would just be the tangent and secant of the complement of θ, so you can imagine a line segment perpendicular to OQ and the lengths of the associated tangent and secant lines from that. https://www.khanacademy.org/cs/why-is-it-called-tangent-and-secant/1269121217 (31 votes) Gabriel Clark 11 years agoPosted 11 years ago. Direct link to Gabriel Clark's post “What about problems like ...” What about problems like (2y-x)/(y^2-3)=5 • (2 votes) Ethan Dlugie 11 years agoPosted 11 years ago. Direct link to Ethan Dlugie's post “Well, you could either do...” Well, you could either do quotient rule or cross multiply, which is easier. (24 votes) Varun Raghavendra 10 years agoPosted 10 years ago. Direct link to Varun Raghavendra's post “How can you consider y as...” How can you consider y as a function of x? Because for every x, there are two y s. Isn't this contradictory to the definition of a function? • (10 votes) Gerhard 9 years agoPosted 9 years ago. Direct link to Gerhard's post “You are right y is _not_ ...” You are right y is not a function of x. You have to define 2 functions to describe a circle with functions. The video explains this from (1 vote) Aztec Binaynay 8 years agoPosted 8 years ago. Direct link to Aztec Binaynay's post “Sorry didn't start my cal...” Sorry didn't start my calculus here in Khanacademy so I don't really know what the derivative notation Sal is using here... What does it mean to have derivative of something with respect to something? • (7 votes) J E 8 years agoPosted 8 years ago. Direct link to J E's post “dy/dx is basically anothe...” dy/dx is basically another way of writing y' and is used a lot in integral calculus. dy/dx is said to be taking the derivative of y with respect to x (sort of like 'solve for y in terms of x' - type terminology). So dy/dt would be taking the derivative of y with respect to t where t is your independent variable. (12 votes) Jason Pickles 10 years agoPosted 10 years ago. Direct link to Jason Pickles's post “at about 4:30, when Sal i...” at about • (7 votes) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “He's using both. The chai...” He's using both. The chain rule is required here because we're differentiating with respect to x and the expression includes y. We're able to do this by treating y as a function of x. This means that y² is actually a composition of two functions: the squaring function applies to whatever function would turn x into y. We don't necessarily know what that function is, but that's okay, it's why we're doing implicit differentiation instead of normal (explicit) differentiation. (9 votes)Want to join the conversation?
Consider the function f value of which is dependent on value of x and y. For example f(x,y) = x + 3 * y where value of f(x,y) is dependent on values of x and y; however x or y are not necessarily dependent on each other.In such a function you can have a general df/dx and df/dy but you can't define a general dy/dx unless you define it based on a specific value of f(x,y).
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/index.htm
Maybe I don't fully understand WHY he does this operation...y=x^2 and I asked, "What is the derivative?" you could easily tell me that dy/dx=2x. Notice, you took the derivative wrt. x of both sides: d/dx(y)=d/dx(x^2) -> dy/dx=2x
Sal is allowed to solve for dy/dx as he does thanks to the chain rule. If I said 2y-2x=1 and I said find the derivative wrt. x, you would think that it is easy. Solve for y and take the derivative: dy/dx=1.
Now I say, "take the derivative before solving for y". Alright:d/dx(2y-2x)=d/dx(1) -> 2*dy/dx-2=0 -> dy/dx=1. The reason that I could just continue with the notation "dy/dx" is because y is a function of x, but I don't know what exactly its relationship to x is. Therefore, I leave dy/dx as an abstract quantity.
The real use of implicit differentiation is when you can't just solve for x. Say I have something weird like cos(x*y)=sin(x). I don't know about you, but I'm not sure how to solve in terms of x for that guy! So, we take the derivative:d/dx cos(x*y) = d/dx sin(x)
dcos(x*y)/d(x*y) * d(x*y)/dx = cosx (I used chain rule on the left side)
-sin(x*y) * (x*dy/dx+y*1) = cosx (I used product rule)
x*dy/dx+y = -cosx/sin(x*y)
dy/dx = ( -cosx/sin(x*y) - y) / x
It's not pretty, but it sure works! The only setback with this is that the derivative is now in terms of both x and y. So, instead of just plugging in values of x, we have to plug in values of x and y (i.e. a coordinate on the original graph) to find the derivative at a point.
Hope that helps!
3:08 4:51
d/dx(x²) = 2(x) (dx/dx) = 2x
Of course, dx/dx = 1 and is trivial, so we don't usually bother with it.
d/dx (y²) = d(y²)/dy (dy/dx) = 2y dy/dx
(Remember that x' = 1 and is almost always omitted, but I will write it just for clarity)
x² + y² = 1
2x(x') + 2y(y') = 0
x(x') + y (y') = 0 ← I just divided by 2
y(y') = −x(x')
y' = −x(x') / y
y' = -x/y
(Of course, we wouldn't really bother writing x' since it is trivially equal to 1.)
d/dx y = d/dx x²
dy/dx = 2x
I seem to be able to do the problems where I am using the product rule but when I get one of these I just cannot seem to get the correct answer. Can't find any examples anywhere for these types of problems.
2y-x=5y^2-15
2*dy/dx - 1 = 10y*dy/dx
2*dy/dx-10y*dy/dx=1
dy/dx*(2-10y) = 1
dy/dx = 1/(2-10y) 0:20
Hope this helps! 4:30
Video transcript
So we've got theequation x squared plus y squared is equal to 1. I guess we could callit a relationship. And if we were to graphall of the points x and y that satisfiedthis relationship, we get a unit circle like this. And what I'm curiousabout in this video is how we can figure out theslope of the tangent line at any point ofthis unit circle. And what immediately might bejumping out in your brain is, well a circle defined thisway, this isn't a function. It's not y explicitlydefined as a function of x. For any x value we actuallyhave two possible y's that satisfy thisrelationship right over here. So you might be temptedto maybe split this up into two separatefunctions of x. You could say y is equal tothe positive square root of 1 minus x squared. And you could say y is equal tothe negative square root of 1 minus x squared. Take the derivatives ofeach of these separately. And you would be able to findthe derivative for any x, or the derivative of theslope of the tangent line at any point. But what I want todo in this video is literally leveragethe chain rule to take thederivative implicitly. So that I don't haveto explicitly define y is a function of x either way. And the way we dothat is literally just apply the derivative operatorto both sides of this equation. And then apply what weknow about the chain rule. Because we are notexplicitly defining y as a function of x, andexplicitly getting y is equal to f primeof x, they call this-- which is really just anapplication of the chain rule-- we call it implicitdifferentiation. And what I want you to keepin the back of your mind the entire timeis that it's just an application ofthe chain rule. So let's apply the derivativeoperator to both sides of this. So it's the derivativewith respect to x of x squared plus ysquared, on the left hand side of our equation. And then that's going tobe equal to the derivative with respect to x onthe right hand side. I'm just doing thesame exact thing to both sides of this equation. Now if I take the derivativeof the sum of two terms, that's the same thing as takingthe sum of the derivative. So this is going tobe the same thing as the derivative withrespect to x of x squared, plus the derivative withrespect to x of y squared. I'm writing all myorange stuff first. So let's see. This is going to be x squared,this is going to be y squared. And then this is going tobe equal to the derivative with respect to x of a constant. This isn't changingwith respect to x. So we just get 0. Now this first termright over here, we have done many, many,many, many, many, many times. The derivative withrespect to x of x squared is just the power rule here. It's going to be 2 timesx to the first power. We could just say 2x. Now what's interesting is whatwe're doing right over here. The derivative withrespect to x of y squared. And the realization here isto just apply the chain rule. If we're taking thederivative with respect to x of thissomething, we just have to take the derivative--let me make it clear-- we're just going to take thederivative of our something. The derivative of y squared--that's what we're taking, you can kind of view that asa function-- with respect to y and then multiply thattimes the derivative of y with respect to x. We're assuming that y doeschange with respect to x. y is not some type ofa constant that we're writing just an abstract terms. So we're taking thederivative of this whole thing with respect to y. Once again, just the chain rule. And then we're takingthe derivative of y with respect to x. It might be a little bitclearer if you kind of thought of it as the derivativewith respect to x of y, as a function of x. This might be, or y isa function of x squared, which is essentially another wayof writing what we have here. This might be alittle bit clearer in terms of the chain rule. The derivative ofy is a function of x squared withrespect to y of x. So the derivative ofsomething squared with respect to that something, times thederivative of that something, with respect to x. This is just the chain rule. I want to say itover and over again. This is just the chain rule. So let's do that. What do we get on theright hand side over here? And I'll write itover here as well. This would be equal to thederivative of y squared with respect to y, is justgoing to be 2 times y. 2 times y, just anapplication of the chain rule. And the derivative ofy with respect to x? Well, we don'tknow what that is. So we're just going to leavethat as times the derivative of y with respect to x. So let's just writethis down over here. So we have is 2x plus thederivative of something squared, with respect tothat something, is 2 times the something. In this case, the somethingis y, so 2 times y. And then times the derivativeof y with respect to x. And this is all goingto be equal to 0. Now that was interesting. Now we have an equationthat has the derivative of a y with respect to x in it. And this is what weessentially want to solve for. This is the slope of thetangent line at any point. So all we have to doat this point is solve for the derivative ofy with respect to x. Solve this equation. So let's do that. And actually just so wecan do this whole thing on the same page so we can seewhere we started, let me copy and paste this up here. This is where we left off. And let's continue there. So let's say let's subtract2x from both sides. So we're left with 2ytimes the derivative of y, with respect to x, is equalto-- we're subtracting 2x from both sides-- so it'sequal to negative 2x. And if we really want tosolve for the derivative of y with respect to x, we canjust divide both sides by 2y. And we're left withthe derivative of y with respect to x. Let's scroll down a little bit. The derivative ofy with respect to x is equal to, wellthe 2s cancel out. We we're left withnegative x over y. So this is interesting. We didn't have to usexplicitly define y as a function of x here. But we got our derivativein terms of an x and a y. Not just only in terms of an x. But what does this mean? Well, if we wantedto find, let's say we wanted to find thederivative at this point right over here. Which, if you're familiarwith the unit circle, so if this was a45 degree angle, this would be the squareroot of 2 over 2 comma the square root of 2 over 2. What is the slope ofthe tangent line there? Well, we figured it out. It's going to benegative x over y. So the slope of thetangent line here is going to beequal to negative x. So negative square rootof 2 over 2 over y. Over square root of 2 over 2,which is equal to negative 1. And that looks just about right.
