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.

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.

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.

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

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].

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

Pretty much, yeah.

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]

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

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

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

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

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.

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, ...?

Argh! Too many brackets

 

The idea is right though