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Posted

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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Posted

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.

Posted

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.

Posted

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].

Posted
: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]

Posted

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.

Posted

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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