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NoahsArk

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NoahsArk

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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}##.

[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.

The figure here is circle. The equation is rewritten in the explicit y=f(x) form asWhat are we really asking when we are asking what the derivative of y2 is?

[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

Mark44

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A shorter version than in the previous post is that when you differentiate ##x^2##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?

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}##.

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NoahsArk

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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. What I think we are asking is to find the instaneous change inA shorter version than in the previous post is that when you differentiate x2with respect to x

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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: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 inywith 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 ##?

$$x^2 + f(x)^2 = 1$$

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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Stephen Tashi

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You mean "why aren't webut we aren't adding an ## x\prime ## to the 2x?

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.

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##.What I think we are asking is to find the instaneous change inywith respect to x at any given point on the curve

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.

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##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: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 ##?

$$\text{Let} \ f(x) = \sin x$$

or, whatever. That would get very tiresome.

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NoahsArk

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Mark44

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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##

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.

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NoahsArk

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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 ##The derivative of x2with respect to x, is 2x. The notation ddx(x2) indicates that this derivative is with respect to x.

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If ##x## is a function of time, then so is ##x^2##. A good example is Kinetic Energy: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

$$E = \frac 1 2 m v(t)^2$$ $$ \frac{dE}{dt} = mv(t)\frac{dv(t)}{dt}$$

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Mark44

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No. The symbol ## \frac {d} {dx} ## is anI 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} ## ?

The symbol ## \frac {dy} {dx} ## is the

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.

Sort of. The derivative of ##x^2##NoahsArk said:Also, and this shows that I still have some difficulty with the basic calculus concepts, what does "the derivate of x^2with 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?

We would have to assume that x is some function of another variable, say t.NoahsArk said:What would it even mean to say the derivative of ## x^2 ## with respect to some variable other than x? Thanks

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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mathwonk

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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

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