Freeman Posted December 7, 2004 Share Posted December 7, 2004 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?! Link to comment Share on other sites More sharing options...
Babbler Posted December 7, 2004 Share Posted December 7, 2004 The formulas are good. Link to comment Share on other sites More sharing options...
fuhrerkeebs Posted December 7, 2004 Share Posted December 7, 2004 Actually the integral would be y=mxn+1/(n+1)+b Link to comment Share on other sites More sharing options...
MandrakeRoot Posted December 7, 2004 Share Posted December 7, 2004 I think with integral you mean primitive here ! Mandrake Link to comment Share on other sites More sharing options...
Dave Posted December 7, 2004 Share Posted December 7, 2004 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]. Link to comment Share on other sites More sharing options...
matt grime Posted December 7, 2004 Share Posted December 7, 2004 They arent' differentials either. Differentials are something else beyond the topic of this question. Link to comment Share on other sites More sharing options...
MandrakeRoot Posted December 8, 2004 Share Posted December 8, 2004 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 Link to comment Share on other sites More sharing options...
CPL.Luke Posted December 24, 2004 Share Posted December 24, 2004 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)) Link to comment Share on other sites More sharing options...
bloodhound Posted December 24, 2004 Share Posted December 24, 2004 if y =mx + b , then y'=m y'=0 iff y = constant. Link to comment Share on other sites More sharing options...
JaKiri Posted December 24, 2004 Share Posted December 24, 2004 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. Link to comment Share on other sites More sharing options...
CPL.Luke Posted December 24, 2004 Share Posted December 24, 2004 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. Link to comment Share on other sites More sharing options...
Dave Posted December 25, 2004 Share Posted December 25, 2004 Rate of change of what? I'm confused. Link to comment Share on other sites More sharing options...
CPL.Luke Posted December 25, 2004 Share Posted December 25, 2004 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. Link to comment Share on other sites More sharing options...
Dave Posted December 25, 2004 Share Posted December 25, 2004 Well, they do represent rates of change.. it's just I'm a little confused to what the question actually is. Link to comment Share on other sites More sharing options...
CPL.Luke Posted December 25, 2004 Share Posted December 25, 2004 that you could actually take a derivative of y=mx+b it just wouldn't be of use Link to comment Share on other sites More sharing options...
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