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Implicit differentiation

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)
Some relationships cannot be represented by an explicit function. For example, x²+y²=1. Implicit differentiation helps us find ​dy/dx even for relationships like that. This is done using the chain ​rule, and viewing y as an implicit function of x. For example, according to the chain rule, the derivative of y² would be 2y⋅(dy/dx). Created by Sal Khan.

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• Would this be considered multivariable calculus?
• 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.
• 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?
• Actually it is an interesting question. Let's start with an example of non-related variables:
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).

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:
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/index.htm
• 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)?
Maybe I don't fully understand WHY he does this operation...
• 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 `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!
• 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...
• 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
d/dx y = d/dx x²
dy/dx = 2x
• From to , Sal tries to explain that d/dx of y^2 is the same as d/dx of y(x)^2 and that this is an application of the chain rule. I am missing how y(x)^2 is a compound function for which the chain rule applies. What is the inside function and what is the outside function? Is dy^2/dy for the outside function?
• 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:
d/dx(x²) = 2(x) (dx/dx) = 2x
Of course, dx/dx = 1 and is trivial, so we don't usually bother with it.

We do the same thing with y², only this time we won't get a trivial chain rule
d/dx (y²) = d(y²)/dy (dy/dx) = 2y dy/dx

Personally, I think using the y' and x' notation is easier for implicit differentiation. Here is the same computation using that notation:
(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.)
• 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?
• 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.

• What about problems like (2y-x)/(y^2-3)=5
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.
(1 vote)
• Well, you could either do quotient rule or cross multiply, which is easier.
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)
• 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?
• 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 within the following 50 seconds.