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?!

The formulas are good.

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

I think with integral you mean primitive here !

 

Mandrake

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].

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

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

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))

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

 

y'=0 iff y = constant.

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.

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.

Rate of change of what? I'm confused.

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.

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

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