Confusion on Implicit Differentiation

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  • #1
NoahsArk
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I am confused about implicit differenciation in a few ways. The main confusion is why, in the equation ## x^2 + y^2 = 1 ##, when we are taking the derivative of the left side, ## 2x + 2yy\prime ##, are we adding a ## y\prime ## to the 2y but we aren't adding an ## x\prime ## to the 2x? I also don't even understand what it means to be taking the derivative of the ## y^2 ## portion of this equation. In a regular differentiation example, like finding the derivative of the function ## y = x^2 ##, we are asking what the slope of the curve is on the graph of that function for any given x value. In all the examples I've seen, the derivative is expressed in terms of x. What are we really asking when we are asking what the derivative of ## y^2 ## is? Are we assuming that now x is a function of y instead of y being a function of x like it is in normal differentiation? E.g. in the function ## x = y^2 ## the derivative of this function would be 2y. Thanks
 
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  • #2
anuttarasammyak
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2yy' is product of ##\frac{dy^2}{dy}=2y## and ##\frac{dy}{dx}=y'##. Here ' means ##\frac{d}{dx}##. ##x'=\frac{dx}{dx}=1##. We do not have to multiply it to 2x which comes from ##\frac{dx^2}{dx}##.
What are we really asking when we are asking what the derivative of y2 is?
The figure here is circle. The equation is rewritten in the explicit y=f(x) form as
[tex]y=\pm \sqrt{1-x^2}[/tex]
and calculate y'=dy/dx=f'(x) to get a of tangent line y=ax+b. Another easier way to get it is leave the equation as it is
[tex]x^2+y^2-1=0[/tex]
, differentiate it by x
[tex]\frac{d}{dx}(x^2+y^2-1)=0[/tex]
[tex]2x+2y\frac{dy}{dx}=0[/tex]
and calculate to get dy/dx as a function of point (x, y) on the figured circle.
 
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  • #3
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The main confusion is why, in the equation ## x^2 + y^2 = 1 ##, when we are taking the derivative of the left side, ## 2x + 2yy\prime ##, are we adding a ## y\prime ## to the 2y but we aren't adding an ## x\prime ## to the 2x?
A shorter version than in the previous post is that when you differentiate ##x^2## with respect to x, it's just 2x, but when you differentiate ##y^2## with respect to x, you have to use the chain rule.

IOW, ##\frac d{dx} x^2 = 2x##, but ##\frac d {dx}y^2 = \frac d {dy}y^2 \cdot \frac {dy}{dx} = 2y \cdot \frac {dy}{dx}##.
 
  • #4
NoahsArk
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I think my confusion is more fundamental and is related to the idea of derivatives themselves and what we are trying to do in general with implicit differentiation. Taking the derivative of any term implies there is a function. E.g. asking the question what's the derivative of ## x^2 ## implies that there is a function, ## y = x^2 ##. So when we ask what's the dertivative of ## y^2 ##, what's the function that we are taking the derivative of?

A shorter version than in the previous post is that when you differentiate x2 with respect to x
I am a bit confused with the terminology. When we ask to differentiate ## x^2 ##, I don't think of it like we are asking to differentiate with respect to x. What I think we are asking is to find the instaneous change in y with respect to x at any given point on the curve. Or when you said "differentiate ## x^2 ##", does that implicitly mean "differentiate the function ## y = x^2 ##?
 
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PeroK
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I think my confusion is more fundamental and is related to the idea of derivatives themselves and what we are trying to do in general with implicit differentiation. Taking the derivative of any term implies there is a function. E.g. asking the question what's the derivative of ## x^2 ## implies that there is a function, ## y = x^2 ##. So when we ask what's the dertivative of ## y^2 ##, what's the function that we are taking the derivative of?



I am a bit confused with the terminology. When we ask to differentiate ## x^2 ##, I don't think of it like we are asking to differentiate with respect to x. What I think we are asking is to find the instaneous change in y with respect to x at any given point on the curve. Or when you said "differentiate ## x^2 ##", does that implicitly mean "differentiate the function ## y = x^2 ##?
In order to do implicit differentiation you have to imagine that ##y## can be expressed as a function of ##x##: as in ##y = f(x)##. Now, you have some equation like:
$$x^2 + f(x)^2 = 1$$
Now, you could go ahead and solve that equation to express ##f(x)## explicitly in terms of ##x##:
$$f(x) = \sqrt{1 - x^2}$$
Assuming we want ##y = f(x)## to be positive here. Then we have:
$$f'(x) = \frac{-x}{\sqrt{1 - x^2}}$$
However, sometimes this is not very easy to do. So, if we go back to our equation, we can differentiate both sides:
$$2x + 2f(x)f'(x) = 0$$
And, this is now an implicit equation involving ##f(x)## and ##f'(x)##. To show where this might be useful, imagine we want ##f'(x) = 0##. That gives ##x = 0## directly. We never had to solve for ##f(x)##. The implicit equation was enough to find the turning point at ##x = 0##.
 
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  • #6
Stephen Tashi
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but we aren't adding an ## x\prime ## to the 2x?
You mean "why aren't we multiplying ##2x## by ##x'##". :smile:

One way to look at it is that the derivative of the function ##f(x) = x ## with respect to ##x## is the constant function ##1##. So ##x'## is 1 and we don't bother writing it as a factor.

What I think we are asking is to find the instaneous change in y with respect to x at any given point on the curve
That is the eventual goal. However the process of implicit differentiation begins with taking the derivative of both sides of the equation with respect to ##x##.

The first step in solving a problem may not solve it. For example, in a geometry problem, the goal may be to find the length of segment AB, but to do that we might begin by finding the length of segments AC and CB.
 
  • #7
PeroK
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I am a bit confused with the terminology. When we ask to differentiate ## x^2 ##, I don't think of it like we are asking to differentiate with respect to x. Or when you said "differentiate ## x^2 ##", does that implicitly mean "differentiate the function ## y = x^2 ##?
##x^2## is a function of ##x##. In the same way that ##\sin x## and ##e^x## are functions of ##x##. You don't always have to write:
$$\text{Let} \ f(x) = \sin x$$
or, whatever. That would get very tiresome.
 
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  • #8
NoahsArk
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Sorry I am still confused. I just want to zoom in on a part of the process for now. When we ask what's the dertivative of ## y^2 ##, what's the function that we are taking the derivative of?
 
  • #9
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Sorry I am still confused. I just want to zoom in on a part of the process for now. When we ask what's the dertivative of ## y^2 ##, what's the function that we are taking the derivative of?
It doesn't make sense to say "what's the derivative of ##y^2##?" A derivative of a function is always with respect to some variable.
The derivative of ##x^2## with respect to x, is ##2x##. The notation ##\frac d {dx}(x^2)## indicates that this derivative is with respect to x. This is indicated by the ##dx## in the differentiation operator ##\frac d {dx}##.
To get the derivative of ##y^2## with respect to x, (we're assuming that y represents some differentiable function of x), we need to use the chain rule, as I mentioned back in post #3.

You can't do implicit differentiation if you don't understand the role of the chain rule in it.
 
  • #10
NoahsArk
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The derivative of x2 with respect to x, is 2x. The notation ddx(x2) indicates that this derivative is with respect to x.
I guess part of the confusion is just terminology and the fact that there are so many symbols that represent a derivative. Is ## \frac {d} {dx} ## the same thing as ## \frac {dy} {dx} ## ? Also, and this shows that I still have some difficulty with the basic calculus concepts, what does "the derivate of ## x^2 ## with respect to x"mean? Does the "with respect to x" part mean that, in order to find the slope of a curve, we need to know what the x coordinate is? What would it even mean to say the derivative of ## x^2 ## with respect to some variable other than x? Thanks
 
  • #11
PeroK
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I guess part of the confusion is just terminology and the fact that there are so many symbols that represent a derivative. Is ## \frac {d} {dx} ## the same thing as ## \frac {dy} {dx} ## ? Also, and this shows that I still have some difficulty with the basic calculus concepts, what does "the derivate of ## x^2 ## with respect to x"mean? Does the "with respect to x" part mean that, in order to find the slope of a curve, we need to know what the x coordinate is? What would it even mean to say the derivative of ## x^2 ## with respect to some variable other than x? Thanks
If ##x## is a function of time, then so is ##x^2##. A good example is Kinetic Energy:
$$E = \frac 1 2 m v(t)^2$$ $$ \frac{dE}{dt} = mv(t)\frac{dv(t)}{dt}$$
 
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I guess part of the confusion is just terminology and the fact that there are so many symbols that represent a derivatie. Is ## \frac {d} {dx} ## the same thing as ## \frac {dy} {dx} ## ?
No. The symbol ## \frac {d} {dx} ## is an operator that is applied to a function. It means "take the derivative of whatever function is to the right of it, with respect to x."
The symbol ## \frac {dy} {dx} ## is the derivative of y (considered to be a function of x), taken with respect to the independent variable x.

It's a little like the difference between ##\sin## and ##\sin(\pi/6)##. The first is the name of one of the trig functions, but it doesn't represent a number, because we haven't said what we are to take the sin of. The second is a number, and is equal to 1/2.
NoahsArk said:
Also, and this shows that I still have some difficulty with the basic calculus concepts, what does "the derivate of x^2 with respect to x" mean? Does the "with respect to x" part mean that, in order to find the slope of a curve, we need to know what the x coordinate is?
Sort of. The derivative of ##x^2## with respect to x means that we want to know how ##x^2## changes relative to a change in x.
NoahsArk said:
What would it even mean to say the derivative of ## x^2 ## with respect to some variable other than x? Thanks
We would have to assume that x is some function of another variable, say t.
Then the derivative of ##x^2## with respect to t would be written as ##\frac d{dt}(x^2)##. Using the chain rule, this would be ##\frac d{dt}(x^2) = \frac d{dx}(x^2) \cdot \frac {dx}{dt} = 2x \cdot \frac {dx}{dt} ##.

For example, if ##x(t) = t^3##, then ##\frac d{dt}(x^2) = \frac d{dx}(x^2) \cdot \frac {dx}{dt} = 2x \cdot 3t^2##. If we substitute ##t^3## in for x, then the last expression becomes ##2(t^3)3t^2 = 6t^5##.

If we had just noted that ##(x(t))^2 = t^6##, then ##\frac d{dt}(x^2) = \frac d{dt}(t^6) = 6t^5## as before.
 
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  • #13
NoahsArk
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@Mark44 this is starting to clarify some of my confusion.
 
  • #14
NoahsArk
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So when we differentiate ## y^2 ##, we need to use the chain rule because y is a function of x, but when we differentiate ## x^2 ##, we don't need the chain rule because x is not a function of another variable?
 
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  • #15
mathwonk
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you can use the chain rule if you want, but when ' means derivative with respect to x, then x' = 1, so writing it does not add anything. i.e. the derivative of x^2 is 2xx', but that equals 2x.
 
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It is implicitly understood that y = y(x) unless otherwise specified. Sometimes you will have that x = x(t) and y = y(t) and you have to multiply with x' as well.
 
  • #17
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I am confused about implicit differenciation in a few ways. The main confusion is why, in the equation ## x^2 + y^2 = 1 ##, when we are taking the derivative of the left side, ## 2x + 2yy\prime ##, are we adding a ## y\prime ## to the 2y but we aren't adding an ## x\prime ## to the 2x? I also don't even understand what it means to be taking the derivative of the ## y^2 ## portion of this equation. In a regular differentiation example, like finding the derivative of the function ## y = x^2 ##, we are asking what the slope of the curve is on the graph of that function for any given x value. In all the examples I've seen, the derivative is expressed in terms of x. What are we really asking when we are asking what the derivative of ## y^2 ## is? Are we assuming that now x is a function of y instead of y being a function of x like it is in normal differentiation? E.g. in the function ## x = y^2 ## the derivative of this function would be 2y. Thanks
We are taking the derivative of both sides, the derivative of a sum of two functions is the sum of their derivatives. d(x^2)/dx=2x
y is assumed to be a function of x.
The derivative of y^2 would have been 2y if we were taking the derivative with respect y (its like graphing y^2 vs y and finding the slope of tangent line at the point of interest). Instead we are taking the derivative with respect to x. By the chain rule, as x changes y changes, as a result y^2 changes you can view this composition by the following
Δ y^2/ Δ x= Δ y^2/ Δ y*( Δ y/ Δ x) d(y^2)/dx
 

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