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September 8th, 2017, 04:27 AM   #1
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How to see when to use which rule?

I managed to get through this exercise, because I knew the answer, and I knew that I had to use the chain rule ( ? )
But I still have trouble understanding, how can I know which formula to use? You can take this exercise as an example. Could someone explain, why I had to use the chain rule?

Last edited by skipjack; September 8th, 2017 at 07:10 AM.
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September 8th, 2017, 04:41 AM   #2
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What exercise?

In general, there is no unique way to get an answer. At this stage in your education much of what you learn can be though of as tools in your mathematical toolbox.

Sometimes it's obvious that you should use a hammer or a screwdriver. But sometimes more than one tool looks suitable. Pliers or spanner or wrench? In this case, you pick one up and try it. If it doesn't work, try a different one. Often you can use either the wrench or the spanner and get the same result.

There's no danger of breaking mathematics if you use the wrong tool, and you can always start again. If you do that, you sometimes find that the spanner is better for the job, making it easier, even though the wrench gets the same result.

You just have to try. Real mathematicians don't get the answers first time either. They spend days or weeks trying to puzzle out problems. There's no reason why you should get everything right first time if they can't.
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September 8th, 2017, 04:58 AM   #3
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I already answered on your previous question.
The chain rule is used when your function is more complicated than just x. Because when you have just x then x'=1. But if you have let's say sqrt(x^2+x^3+5) then you have to use the chain rule. Or if you had cos(x^2) or ln (x^2) etc...
I learnt derivatives using this calculator, it is step by step, it may help you:
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September 8th, 2017, 06:39 AM   #4
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A simple example of what I was talking about is the following:
$$\frac{\mathrm d}{\mathrm dx} (2x + 1)^2$$
There are at least three ways to calculate this.
  1. The chain rule:
    Let $u=2x+1$ so that $y = (2x+1)^2 = u^2$. Now the chain rule says that $$\frac{\mathrm dy}{\mathrm dx} = \frac{\mathrm dy}{\mathrm du} \frac{\mathrm du}{\mathrm dx}$$ which gives us \begin{align*}\frac{\mathrm dy}{\mathrm dx} &= 2u \cdot 2 \\ &= 4u \\ \frac{\mathrm d}{\mathrm dx}(2x+1)^2 &= 4(2x+1) \end{align*}
  2. The product rule:
    We notice that $(2x+1)^2 = (2x+1)(2x+1)$ which is a product of two functions of $x$. We let $u=(2x+1)$ and $v=2x+1$. The product rule says that
    $$\frac{\mathrm d}{\mathrm dx}uv = v\frac{\mathrm du}{\mathrm dv} + u\frac{\mathrm dv}{\mathrm dx}$$
    which gives us
    \frac{\mathrm d}{\mathrm dx}uv &= (2x+1)\cdot 2 + (2x+1)\cdot 2 \\
    &= 2(2x+1)+2(2x+1) \\
    \frac{\mathrm d}{\mathrm dx}(2x+1)(2x+1) &= 4(2x+1)
  3. Simplification:
    We can also simplify the function before differentiating. $(2x+1)^2 = 4x^2+4x+1$ and then
    \begin{align*} \frac{\mathrm d}{\mathrm dx}(4x^2+4x+1) &= 8x + 4 \\ &= 4(2x+1)\end{align*}
All of these methods are perfectly valid. Most people would probably say that they prefer the first or the third. I'd guess that most would prefer the first one, but that's probably for the same reason that us boys would prefer to use a power-saw than a manual one.
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September 8th, 2017, 08:03 AM   #5
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$y = 3\sqrt{x} - \dfrac{\small81}{x} + 13 = 3x^{1/2} - 81x^{-1} + 13\implies y^\prime = {\small\dfrac32}x^{-1/2} + 81x^{-2}$,
so $y^\prime = {\small\dfrac12} + 1 = \small\dfrac32$ when $x = 9$.
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September 8th, 2017, 08:26 AM   #6
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Originally Posted by vizzy22 View Post

Could someone explain, why I had to use the chain rule?
Not really, no.

I guess $3\sqrt{x}$ can use the chain rule when $u=\sqrt{x}=x^{\frac12}$ and $y=3u$, but you should quickly get the point (through practice) that multiplicative constants don't require you to resort to particular rules.

Skipjack's solution is the way that I would do it.
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September 8th, 2017, 09:23 AM   #7
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The working below has a few "tricks" that you should familiarize yourself with (if you aren't already familiar with them).

$\displaystyle f(x)=\sqrt x$

$\displaystyle \begin{align*}f'(x)&=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h} \\
&=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}\cdot\frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x} \\
&=\lim_{h\to0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt x)} \\
&=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt x)} \\
&=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt x} \\
&=\frac{1}{2\sqrt x}\end{align*}$
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