# 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:

$\frac{d}{dy}©=0$

$\frac{d}{dy}(x)=1$

$\frac{d}{dy}(u+v+...)=\frac{d}{dy}(u)+\frac{d}{dy}(v)+...$

$\frac{d}{dy}(cu)=c\frac{d}{dy}(u)$

$\frac{d}{dy}(uv)=u\frac{d}{dy}(v)+v\frac{d}{dy}(u)$

$\frac{d}{dy}(uvw)=uv\frac{d}{dy}(w)+uw\frac{d}{dy}(v)+vw\frac{d}{dy}(u)$

$\frac{d}{dy}(\frac{u}{c})=\frac{1}{c}\frac{d}{dy}(u),c\not=0$

$\frac{d}{dy}(\frac{c}{u})=c\frac{d}{dy}(\frac{1}{u})=-\frac{c}{u^2}\frac{d}{dy}(u),u\not=0$

$\frac{d}{dy}(\frac{u}{v})=\frac{v\frac{d}{dy}(u)-u\frac{d}{dy}(v)}{v^2},v\not=0$

$\frac{d}{dy}(x^m)=mx^{m-1}$

$\frac{d}{dy}(u^m)=mu^{m-1}\frac{d}{dy}(u)$

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 $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$. So lets suppose you've got the function $e^{x^2}$.

The first step is to let $u = x^2$. Then we have that $y = e^u$. Hence:

$\frac{dy}{du} = e^u$

$\frac{du}{dx} = 2x$

So $\frac{dy}{dx} = 2x e^u = 2x e^{x^2}$.

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

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Pretty much, yeah.

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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']

$A=LW$ and $2L+2W=300$

$W=150-L$ so $A=150L-L^2$

$da=150-2L$ The maximum is when da is equal to 0, so

$0=150-2L$ $L=75$

When you solve for w, you get 75, so the overall area is

$5625 m^2$

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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 $\frac{\partial}{{\partial}t}$ the same as $\frac{d}{dt}$?

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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 $D_x(f(g(h(x))))=f'(g(h(x)))g'(h(x))h'(x)$, and $D_x(f(g(h(i(x)))))=f'(g(h(x)))g'(h(x))h'(i(x))i'(x)$ for four, ...?

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Argh! Too many brackets

The idea is right though

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