ydoaPs Posted March 30, 2005 Share Posted March 30, 2005 i can see the f(g(bla bla bla)) way, but how do you do it the other way. maybe i should just wait until you get to it. edit: it would appear that with logarithms, you do what i said a few posts ago. Link to comment Share on other sites More sharing options...
ydoaPs Posted April 1, 2005 Share Posted April 1, 2005 here are some differentiation rule thingys: if u, v, and w are differentiable functions, and c and m are constants, then: [math]\frac{d}{dy}©=0[/math] [math]\frac{d}{dy}(x)=1[/math] [math]\frac{d}{dy}(u+v+...)=\frac{d}{dy}(u)+\frac{d}{dy}(v)+...[/math] [math]\frac{d}{dy}(cu)=c\frac{d}{dy}(u)[/math] [math]\frac{d}{dy}(uv)=u\frac{d}{dy}(v)+v\frac{d}{dy}(u)[/math] [math]\frac{d}{dy}(uvw)=uv\frac{d}{dy}(w)+uw\frac{d}{dy}(v)+vw\frac{d}{dy}(u)[/math] [math]\frac{d}{dy}(\frac{u}{c})=\frac{1}{c}\frac{d}{dy}(u),c\not=0[/math] [math]\frac{d}{dy}(\frac{c}{u})=c\frac{d}{dy}(\frac{1}{u})=-\frac{c}{u^2}\frac{d}{dy}(u),u\not=0[/math] [math]\frac{d}{dy}(\frac{u}{v})=\frac{v\frac{d}{dy}(u)-u\frac{d}{dy}(v)}{v^2},v\not=0[/math] [math]\frac{d}{dy}(x^m)=mx^{m-1}[/math] [math]\frac{d}{dy}(u^m)=mu^{m-1}\frac{d}{dy}(u)[/math] ha dave, i beat you to it. Link to comment Share on other sites More sharing options...
Dave Posted April 1, 2005 Author Share Posted April 1, 2005 Yeah... great. But seriously, how many people are following this? If it's just the above then I'm not going to bother writing any more. Link to comment Share on other sites More sharing options...
ydoaPs Posted April 1, 2005 Share Posted April 1, 2005 Shawn said he will. (follow it, not post) Link to comment Share on other sites More sharing options...
Dave Posted April 1, 2005 Author Share Posted April 1, 2005 Re: chain rule (on IRC): Chain rule basically says [math]\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}[/math]. So lets suppose you've got the function [math]e^{x^2}[/math]. The first step is to let [math]u = x^2[/math]. Then we have that [math]y = e^u[/math]. Hence: [math]\frac{dy}{du} = e^u[/math] [math]\frac{du}{dx} = 2x[/math] So [math]\frac{dy}{dx} = 2x e^u = 2x e^{x^2}[/math]. Link to comment Share on other sites More sharing options...
ydoaPs Posted April 1, 2005 Share Posted April 1, 2005 ok, i'm just stupid. it is the same thing written a different way Link to comment Share on other sites More sharing options...
Dave Posted April 1, 2005 Author Share Posted April 1, 2005 Pretty much, yeah. Link to comment Share on other sites More sharing options...
Ducky Havok Posted April 2, 2005 Share Posted April 2, 2005 Suppose you want to build a rectangular paddock to contain some sheep. You want to maximize the area of your paddock because you want to fit as many sheep in as possible' date=' and you have 300m of fencing available to you. Find the maximum area.[/quote'] Nobody answered your easier question so I figured I would. [math]A=LW[/math] and [math]2L+2W=300[/math] [math]W=150-L[/math] so [math]A=150L-L^2[/math] [math]da=150-2L [/math] The maximum is when da is equal to 0, so [math]0=150-2L [/math] [math]L=75[/math] When you solve for w, you get 75, so the overall area is [math] 5625 m^2 [/math] Link to comment Share on other sites More sharing options...
ydoaPs Posted April 2, 2005 Share Posted April 2, 2005 i got that answer using simple arithmatic. i divided 300 by for then squared it. Link to comment Share on other sites More sharing options...
Ducky Havok Posted April 2, 2005 Share Posted April 2, 2005 Yes, its easy to do it using arithmatic or common sense. The most area is of course going to be a square. But the point was to use calculus I think Link to comment Share on other sites More sharing options...
ydoaPs Posted April 2, 2005 Share Posted April 2, 2005 yes, but if you are lazy, like me, it is tempting to go the path of least resistance. Link to comment Share on other sites More sharing options...
ydoaPs Posted April 2, 2005 Share Posted April 2, 2005 is [math]\frac{\partial}{{\partial}t}[/math] the same as [math]\frac{d}{dt}[/math]? Link to comment Share on other sites More sharing options...
Dave Posted April 2, 2005 Author Share Posted April 2, 2005 No. The former is called a partial derivatives and deals with function of more than one variable. I'd rather not get into that though. Link to comment Share on other sites More sharing options...
ydoaPs Posted April 4, 2005 Share Posted April 4, 2005 for three functions is the chain rule [math]D_x(f(g(h(x))))=f'(g(h(x)))g'(h(x))h'(x)[/math], and [math]D_x(f(g(h(i(x)))))=f'(g(h(x)))g'(h(x))h'(i(x))i'(x)[/math] for four, ...? Link to comment Share on other sites More sharing options...
Dave Posted April 4, 2005 Author Share Posted April 4, 2005 Argh! Too many brackets The idea is right though Link to comment Share on other sites More sharing options...
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