Differentials and Integrals(uggh!)

[Freeman]

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 [math]y=mX^n + b[/math] it would be: [math]y=(mn)X^{n-1}[/math] for a differential, right? And [math]y=(m/n)X^{n+1} + b[/math] for an integral...right?!

[Babbler]

The formulas are good.

[fuhrerkeebs]

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

[MandrakeRoot]

I think with integral you mean primitive here !

 

Mandrake

[Dave]

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: [math]y=(mn)X^{n-1}[/math] for a differential, right? And [math]y=(m/n)X^{n+1} + b[/math] for an integral...right?!

 

Remember that you need to get your notation right.

 

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

[matt grime]

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

[MandrakeRoot]

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

[CPL.Luke]

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))

[bloodhound]

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

 

y'=0 iff y = constant.

[JaKiri]

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.

[CPL.Luke]

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.

[Dave]

Rate of change of what? I'm confused.

[CPL.Luke]

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.

[Dave]

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

[CPL.Luke]

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