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Dave

Calculus I - Lesson 3: Properties of the derivative

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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.

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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.

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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.

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Shawn said he will. (follow it, not post)

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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].

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ok, i'm just stupid. it is the same thing written a different way

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:P

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]

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i got that answer using simple arithmatic. i divided 300 by for then squared it.

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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

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yes, but if you are lazy, like me, it is tempting to go the path of least resistance.

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is [math]\frac{\partial}{{\partial}t}[/math] the same as [math]\frac{d}{dt}[/math]?

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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.

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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, ...?

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Argh! Too many brackets ;)

 

The idea is right though :)

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