# Differentials and Integrals(uggh!)

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OK, so I have several questions...

First, a differential is a tangent to the curve and the integral is the area between the curve, X axis, and differentials...right?

Second, if I have $y=mX^n + b$ it would be: $y=(mn)X^{n-1}$ for a differential, right? And $y=(m/n)X^{n+1} + b$ for an integral...right?! ##### Share on other sites

The formulas are good.

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Actually the integral would be y=mxn+1/(n+1)+b

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I think with integral you mean primitive here !

Mandrake

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OK' date=' so I have several questions...

First, a differential is a tangent to the curve and the integral is the area between the curve, X axis, and differentials...right?

Second, if I have [math']y=mX^n + b[/math] it would be: $y=(mn)X^{n-1}$ for a differential, right? And $y=(m/n)X^{n+1} + b$ for an integral...right?! Remember that you need to get your notation right.

$y = mx^n + b$ implies that $\tfrac{dy}{dx} = mnx^{n-1}$, but it certainly does not imply that $y = mnx^{n-1}$.

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They arent' differentials either. Differentials are something else beyond the topic of this question.

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Yes indeed, the notation is rather sloppy. It would be better to write y(X), y'(X) and Y(X) for example (and indicating the domain and range of these functions !)

Mandrake

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• 3 weeks later...

for y=mx +b couldn't you take a derivative of it and get y'=m+b this being a constant number it would just tell you that the rate of change is zero

right?

( also shouldn't you never see y=mx^n +b as y=mx+b is a way of writing a linear equation (slope intercept form))

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if y =mx + b , then y'=m

y'=0 iff y = constant.

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for y=mx +b couldn't you take a derivative of it and get y'=m+b this being a constant number it would just tell you that the rate of change is zero

right?

db/dx = 0.

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wups yeah your right it would be y'=m

but the rest of my statement would be right correct?

edit**

I was not trying to say that y'=0 but rather the change in the expression is zero.

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Rate of change of what? I'm confused.

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sorry just my bad wording. the way I learned derivatives was that they represented the rate of change in an expression. its just my way of thinking about it.

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Well, they do represent rates of change.. it's just I'm a little confused to what the question actually is.

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that you could actually take a derivative of y=mx+b it just wouldn't be of use

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