What is the difference between representing different ways of differentiation

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JustHoping
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#1
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Okay, so I can differentiate a function and find the gradient of a graph. I can do partial differentiation, differentiation of several variables... that sort of thing.

But i still have no idea what the difference is between representing the change as
dy/dx

and when d is triangle. Sometimes formualas will use both and they seem to mean different things. Or the first part of the question will use on and the second will use another for some purpose I can't figure out. :/

please help!
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Pangol
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Conventially, the triangle (Greek capital delta) means "a change in", usually taken to possibly mean a large change. So, if you've moved from a point on a curve to a nearby point where x and y have changed by delta x and delta y, the gradient of the line joining the two points is delta y / delta x.

Differentiation is the process of seeing what happens when the amount of change gets progressively smaller and smaller (I'm sure you've seen this sort of limiting process), and it is only in the limit as the amount of change tends to zero that we should use dy/dx.

Maths texts should be fairly hot on the difference, but some physical sciences can play a bit more freely with the terminology.
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pereira325
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Differentiation measures the rate of chang or something similar to that. When you do dy/dx it means differentiating y in respect to x. Same with any other thing in that form, i.e. dy/dt, or dt/dx. Could you give an example of a formula using the triangle?
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Notnek
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(Original post by pereira325)
Differentiation measures the rate of chang or something similar to that. When you do dy/dx it means differentiating y in respect to x. Same with any other thing in that form, i.e. dy/dt, or dt/dx. Could you give an example of a formula using the triangle?
I believe the OP is talking about the difference between \frac{\Delta y}{\Delta x} and \frac{dy}{dx}.
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JustHoping
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(Original post by notnek)
I believe the OP is talking about the difference between \frac{\Delta y}{\Delta x} and \frac{dy}{dx}.
Yes I am thank you! I couldn't seem to get it in the actual question
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atsruser
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(Original post by JustHoping)
Yes I am thank you! I couldn't seem to get it in the actual question
I'm short of time but in brief:

1. \frac{dy}{dx} represents the gradient of a tangent line to a curve at a point P i.e the gradient of the curve at P.

2. \frac{\Delta y}{ \Delta x} represents the gradient of a secant line between two points P and Q on a curve. If we consider calculating the value of \frac{\Delta y}{ \Delta x}as we make Q closer and closer to P, then that values approaches \frac{dy}{dx} more and more closely.

e.g. pick two points P,Q on the curve y=x^2 with P(x,x^2), Q(x+h, (x+h)^2) then by noting that the gradient of the secant PQ is given by rise/run, we have:

\frac{\Delta y}{ \Delta x} = \frac{(x+h)^2 -x^2}{(x+h)-x} = \frac{2xh+h^2}{h} = 2x+h (*)

We can make Q approach P as near as we like by making h smaller, and then we can make the gradient of PQ as close to 2x as we like. This is the idea of finding the limit of the gradient of PQ as h approaches 0. Since the limit as h approaches 0 is 2x, we say that the gradient of the curve at P is 2x i.e.

\displaystyle \frac{dy}{dx} = \lim_{h \to 0} \frac{\Delta y}{ \Delta x} = \lim_{h \to 0} 2x+h = 2x

Note that we can't just say: Oh, we'll just set h=0 to find the gradient then! - that goes wrong, because if we set h=0, then:

a) algebraically the division by h that we did in (*) is undefined, and
b) geometrically, our picture of a triangle with a secant line hypotenuse and run of h and rise of 2xh+h^2 is wrong since if h=0, then P and Q are on top of each other i.e. our triangle on which we were basing our whole argument no longer exists.

3. In scientific experiments, we often measure a finite change in a quantity when we make a finite change to another quantity e.g. we may change the voltage across a device by 0.01 V, and we see a change in the current through it of 0.03 A. Then we can find, approximately, the gradient of the I-V curve at V=0.01 V by calculating \frac{\Delta V}{ \Delta I} = \frac{1}{3} \Omega.

That's why you'll often see this notation in physics or engineering texts. For example, in electronics, the quantity \frac{\Delta V}{ \Delta I} is referred to as the "small signal resistance" - it's a measurable (or calculable) approximation to the gradient of the V-I curve of a device.
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