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I can't understand what the notation means, it doesn't make sense to me, it is too complex for me to think of. Anyways I just can't understand what $f'\bigl(g(x)\bigr)g'(x)$ means, or the crazy $dy/dx = (dy/du)(du/dx)$.

I made sense of it in class that if the problem is $(4x+1)^4$ then I just drop down the 4 and get the derivative at the end to make it $4(4x+1)(4)$ but I don't think that is correct either. I tried to find some videos online but they don't seem to help, they made things infinitely more confusing since they did it differently than what I learned in class.

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    ...from what I've seen with your previous stuff, it looks as if you need to brush up on the algebra first and foremost.2011-09-22
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    In any event, what do you understand about composing functions? Can you see how $(4x+1)^4$ is a composition?2011-09-22
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    Thanks, that is exactly what I wanted to hear about learning the chain rule. Thanks a lot. I know about functions like simples ones like that will be something like v^4 and v=4x+12011-09-22
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    Lastly: it's never a good idea to be on the Internet when you're angry. Calm down, cool yourself, take a break. When you're in a better mood, you can get back to asking questions.2011-09-22
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    I can't take a break, I have to spend all my time from now until monday studying so I can get a D on the test.2011-09-22
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    But back to the chain rule: you had it right, you have the composition of the function $f(x)=x^4$ and the function $g(x)=4x+1$. The chain rule applies like this: differentiate $f(x)$, replace the $x$ with $g(x)$, and then multiply the result of that with the derivative of $g(x)$. Your slip-up was in differentiating $f(x)$...2011-09-22
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    Jordan, have you considered hiring a tutor? You might even be able to find someone willing to work with you *for free*...2011-09-22
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    How did you learn it in class?2011-09-22
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    I don't know what differentiate means.2011-09-22
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    In class we were taught to seperate out the function and then carry it down somehow. I didn't write it down (advice I was given, just pay attention in class and take notes after class) but the book does it differently and isn't in any way helpful at all. I think stewart is some sort of math genius and assumes everyone else is as well so he often omits several steps and just wizards out an answer. SDo when I put in g(x) into f(x) that means what? Do I change f(x) into a simple version like (4x+2)^2 would then be f(x)=x^2 and g(x) is 4x+2 and f(g(x)) is (4x+2)^2?2011-09-22
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    "Differentiate" means "take the derivative of". It's the main thing in the first part of calculus. And your example of composition, while different than the one you originally mentioned, is correct.2011-09-22
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    "...(if) f(x)=x^2 and g(x) is 4x+2 and f(g(x)) is (4x+2)^2?" - right. You're not as bad as you think you are.2011-09-22
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    So my book gives this ridiculous example of $y=tan\pi x$ do I use the product rule first as tanpi and x? I jsut have trouble applying this rule, stewart immediatley jumps to stuff I just don't understand.2011-09-22
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    $\pi x$ is the *argument* (i.e. "inside") of tangent. After taking the derivative of $\tan$, you would then multiply - as per the chain rule - by the derivative of the *inside*... namely $\pi$2011-09-22
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    I don't know what an argument is or how something can be inside of tangent.2011-09-22
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    You must take the tangent of *something*. Have you studied the unit circle? The tangent of $\pi / 4$ is 1; the tangent of $0$ is $0$, etc. In these examples, $\pi / 4$ and $0$ are the arguments - the things of which we are taking the tangent.2011-09-22
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    Yes I got a pretty solid C- in trig, and promptly forgot everything I learned. So we are finding the tanget if pi or the tanget of pix?2011-09-22
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    I don't know what is wrong with me, but I just can't get any of these homework problems right so I am goign to take a break. I am not going to do well on this test anyways so it hardly matters how much I study.2011-09-22
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    The latter. In the ridiculous example, the argument of tangent is $\pi x$. $f(x) = \tan (\pi x)$. Then $f'(x) = \sec^2(\pi x) \cdot \pi$2011-09-22
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1407/discussion-between-the-chaz-and-jordan-carlyon)2011-09-22
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    In what context are you using the chain rule? Usually, it is used to differentiate a composition of functions. To differentiate a function is to take its derivative. Do you know how to take a derivative?2011-09-22

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I'll show you a couple examples of the Chain Rule, where I write out every step I can; hopefully this will help. All you need to know is how to break up a complicated function into pieces (by composition), and how to take the derivative of each piece.

I will take the derivative of the following functions:

  1. $y=(5x^2+4)^3$
  2. $y=\sqrt{\sin x}$
  3. $y=\sec(x^4)$

$1. \quad$ $y=(5x^2+4)^3$ should be viewed as $y=u^3$ where $u=5x^2+4$.

Now, we need to take the derivative of each piece. I get $$\frac{dy}{du} = 3u^2$$ for the derivative of the first piece and $$\frac{du}{dx}=10x$$ for the derivative of the second piece.

The chain rule tells me that my final answer (that is, $\frac{dy}{dx}$) is equal to $\frac{dy}{dx} = (3u^2)(10x)$. But we're not quite done. The variable $u$ is something I made up, so it shouldn't be part of my final answer. Since $u=5x^2+4$, I can just plug this in and get the final answer $$\frac{dy}{dx} = (3(5x^2+4)^2)(10x).$$

$2. \quad$ $y=\sqrt{\sin x}$ should be viewed as $y=\sqrt{u}$ where $u=\sin x$.

I get $$\frac{dy}{du} = \frac{1}{2\sqrt{u}}$$ for the derivative of the first piece and $$\frac{du}{dx}=\cos x$$ for the derivative of the second piece.

Then $\frac{dy}{dx} = (\frac{1}{2\sqrt{u}})(\cos x)$. Since $u=\sin x$, I can just plug this in and get the final answer $$\frac{dy}{dx} = \frac{\cos x}{2\sqrt{\sin x}}.$$

$3. \quad$ $y=\sec(x^4)$ should be viewed as $y=\sec u$ where $u=x^4$.

I get $$\frac{dy}{du} = \sec u \tan u$$ for the derivative of the first piece and $$\frac{du}{dx}=4 x^3$$ for the derivative of the second piece.

So $\frac{dy}{dx} = (\sec u \tan u)(4x^3)$. Now I plug in $u=x^4$to get the final answer $$\frac{dy}{dx} = (\sec x^4 \tan x^4)(4x^3).$$


The notation that confuses you is really a formula instructing you to do each of these steps. For your test, you probably want to practice just getting the right answer to these problems first (and being good enough at it that you can spot mistakes when you make them); but you should spend some time figuring out the formulas $$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$ and $$ (f \circ g )'(x) = f'(g(x))g'(x).$$ These formulas are useful, and being able to read and use them will help you in the future. Reading a mathematical text is a skill to work on patiently; you will find that you improve at it.

It is normal that in the rest of your calculus course, you will not write out every step as I have done here. In fact, you can expect your teacher and your book to take these Chain Rule derivatives in just one step-- the thought process still follows the steps I describe.

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    Only (+1), until I can figure out how to donate my rep to your cause!2011-09-22
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    @Jordan: You should also practice messier examples than the 3 I gave. One common complication is a Chain Rule problem requiring extra steps, like finding the derivative of $f(x) = \sin(\cos(x^2))$.2011-09-22
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    Thanks a lot, I see where I was going wrong before. Instead of just plugging in for u I was just trying to draw out the equation like f(g(x)) or whatevr and that was confusing me.2011-09-22
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    Sorry I thought I understood it but I don't. Don't know what is wrong with me, I know this is really simple but I can't do it. Anyways I am trying to work a problem out of my book $(x^4+3x^2 -2)^5$ I make that into $y=u^5$ and u=$(x^4+3x^2 -2)$ Then I mash it together and get $ 5((x^4+3x^2 -2)^5 (4x^3+6x)$ something is wrong though as my book gets a different answer.2011-09-22
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    @Jordan Carlyon:Is it $5\times (x^4+3x^2 -2)^4 \times (4x^3+6x)$?2011-09-22
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    @Jordan: You're close-- the right answer is $5(x^4+3x^2-2)^4(4x^3+6x)$. (Do you see why?) We could rearrange or simplify the answer a lot of different ways, like $(20x^3+30x)(x^4+3x^2-2)^4$. Maybe that's what the book is doing?2011-09-22
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    Or May be this $10 x \left(3+2 x^2\right) \left(-2+3 x^2+x^4\right)^4$ or even this $480 x-2560 x^3+3600 x^5+1520 x^7-4770 x^9-1140 x^{11}+2100 x^{13}+1280 x^{15}+270 x^{17}+20 x^{19}$? However the later is crazy expansion!2011-09-22
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    The book is getting $10x(x^4+3x^2 -2)^4 (2x^2 +3)$2011-09-22
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    @Jordan: Yes, that's the same as our answers. Write the book's answer side-by-side with $5(x^4+3x^2-2)^4(4x^3+6x)$, and figure out how all the pieces of both answers match.2011-09-22
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    @Jordan Carlyon:I thought so,it is the most simplified form also agreed by [wolfarm alpha](http://www.wolframalpha.com/input/?i=Simplify%5B%2820+x^3+%2B+30+x%29+%28x^4+%2B+3+x^2+-+2%29^4%5D).To get that manually from $5\times (x^4+3x^2 -2)^4 \times (4x^3+6x)$,factor out $2$ from $(4x^3+6x)$ which will give $2\times (2x^3+3)$,and now combining with our rest answer we will get $5x\times (x^4+3x^2 -2)^4 \times 2 \times (2x^2 +3)=10x(x^4+3x^2 -2)^4 (2x^2 +3)$.We are done!2011-09-22
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    Oh I get it now they just pulled out a 2 from the last part.2011-09-22
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If $y$ is changing 7 times as fast as $u$ is changing and $u$ is changing 5 times as fast as $x$ is changing, then $y$ is changing 35 times as fast as $x$ is changing.

That's what the chain rule means.

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If you were to take the chain rule of said function you posted, this is what it would look like.

\begin{equation} (4)(4x + 1)^3(4) \end{equation}

This answer is achieved by looking at the chain rule this way. Whenever you have some function to a power, you take the power and multiply the equation by it and subtract the power by 1. After that, just take the derivative of the actual inside function, which in this case would be 4x + 1 (the derivative would be 4) and multiply by it. Sorry for my poor wording also, haha. If you have any questions, I can try and clear it up some more.

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    Sorry, I don't know how to make the square root symbol yet because I just joined this site, lolz.2011-09-28
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    (1) You can right click > "show source" on others' posted equations to see the LaTeX markup they used. (2) You can test out markup in [codecogs' editor](http://www.codecogs.com/latex/eqneditor.php). (3) You can use [Wikipedia](http://en.wikipedia.org/wiki/Help:Displaying_a_formula) for reference on most of the LaTeX you'll ever need to use.2011-09-28