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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    $\begingroup$ ...from what I've seen with your previous stuff, it looks as if you need to brush up on the algebra first and foremost. $\endgroup$ – J. M. is a poor mathematician Sep 22 '11 at 17:58
  • $\begingroup$ In any event, what do you understand about composing functions? Can you see how $(4x+1)^4$ is a composition? $\endgroup$ – J. M. is a poor mathematician Sep 22 '11 at 17:59
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    $\begingroup$ Jordan, have you considered hiring a tutor? You might even be able to find someone willing to work with you for free... $\endgroup$ – Altar Ego Sep 22 '11 at 18:04
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    $\begingroup$ I don't know what differentiate means. $\endgroup$ – toby yeats Sep 22 '11 at 18:05
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    $\begingroup$ let us continue this discussion in chat $\endgroup$ – The Chaz 2.0 Sep 22 '11 at 18:21

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.

  • $\begingroup$ Only (+1), until I can figure out how to donate my rep to your cause! $\endgroup$ – The Chaz 2.0 Sep 22 '11 at 19:02
  • $\begingroup$ @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 $f(x) = \sin(\cos(x^2))$. $\endgroup$ – Jonas Kibelbek Sep 22 '11 at 19:08
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    $\begingroup$ 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. $\endgroup$ – toby yeats Sep 22 '11 at 19:16
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    $\begingroup$ @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. $\endgroup$ – Jonas Kibelbek Sep 22 '11 at 19:40
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    $\begingroup$ Oh I get it now they just pulled out a 2 from the last part. $\endgroup$ – toby yeats Sep 22 '11 at 19:42

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.


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.

  • $\begingroup$ Sorry, I don't know how to make the square root symbol yet because I just joined this site, lolz. $\endgroup$ – BleuCheese Sep 28 '11 at 3:17
  • $\begingroup$ (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. (3) You can use Wikipedia for reference on most of the LaTeX you'll ever need to use. $\endgroup$ – anon Sep 28 '11 at 3:23

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