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Old 2018-12-15, 02:36   #122
Chuck
 
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Quote:
Originally Posted by jvang View Post
Starting with \[\dfrac{1}{2\sqrt{x}}\]
Rewrite as \[\dfrac{1}{2}x^{-\frac{1}{2}}\]

(I'm not very fluent with this $$ math coding so please excuse).

Now just use the power rule. minus 1/2 times 1/2 and subtract one from the power of x.

Result is -1/(4 x^(3/2))

Last fiddled with by LaurV on 2018-12-15 at 04:16 Reason: fixed the missing parenthesis in the formula - now it shows right :)
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Old 2018-12-15, 02:38   #123
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Where can I read about how to use this $$ math formatting?
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Old 2018-12-15, 02:48   #124
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Quote:
Originally Posted by Chuck View Post
Where can I read about how to use this $$ math formatting?
It’s called Latex! \(\LaTeX\)

Most of the time you can just put plain text formatting in and it’ll look better. But you have to use a bit of Latex code for fractions, roots, and other symbols. Google should be able to help you out, and there are online editors with buttons that insert code for you
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Old 2018-12-15, 04:22   #125
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Quote:
Originally Posted by Chuck View Post
(I'm not very fluent with this $$ math coding so please excuse).
You actually got it right, except that a bracket was missing. I fixed it for you. Also, I used \frac instead of \dfrac to let \Latex control the size of indexes and powers.

A very good place to start is to bookmark few lists like this one. You don't need to learn the algorithms how to construct such symbols, just to use what is constructed already, and you also don't need to learn them by heart, you will remember them as you use them day by day, and when you need new stuff, consult the list.

Some may not work, as we use here a subset of Latex, called Mathjax. Some standard symbols are missing, but some other new are introduced. You can also create your own. Also, here on the forum you don't need to use the double dollar, we have \( and respectively \[ to put formulas in text and respectively on separate line (also mathjax feature).

If you still insist in having a WYSIWYG TeX editor, you can look on the web. Most of them cost money, but few are free and moderate powerful. They can let you edit equations like word, etc, and generate LaTeX code that you can paste on the forum. If you use mac, then Compositor is a very good tool.

Trivia: We remember in the nineties we were programming fractals and chess games in latex, hehe... like an "escape from reality" mechanism... (we were "forced" by our uni to type shitty books and articles of/for our professors - not all of them popular with students, as we were thinking at that age...).

Last fiddled with by LaurV on 2018-12-15 at 04:39
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Old 2018-12-15, 09:37   #126
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Quote:
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Now we need the chain rule to get in further (derivative of outside multiplied by derivative of inside, right?)
Absolutely right.
You accidentally differentiated the outside and then the inside instead of differentiating the outside and then multiplying by the derivative of the inside!
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Old 2018-12-15, 13:25   #127
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Quote:
Originally Posted by Chuck View Post
Where can I read about how to use this $$ math formatting?
The BB Code page in the FAQ. Just click on the highlighted code, and you'll see what it does. In my experience, tex gives the nicest-looking displays.
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Old 2018-12-15, 13:58   #128
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Quote:
Originally Posted by LaurV View Post
A very good place to start is to bookmark few URL="https://oeis.org/wiki/List_of_LaTeX_mathematical_symbols"]lists like this one[/URL]. You don't need to learn the algorithms how to construct such symbols, just to use what is constructed already, and you also don't need to learn them by heart, you will remember them as you use them day by day, and when you need new stuff, consult the list.
Thanks. When I searched on LATEX math formatting I found several good references.
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Old 2018-12-15, 14:56   #129
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Quote:
Originally Posted by Dr Sardonicus View Post
You forgot to apply the chain rule. You have f = 2*sqrt(x) and want the derivative of 1/f.

(1/f)' = (-1)*f^(-2)* f'

It's a lot easier, though, as has already been pointed out, to pull out the original factor 1/2 -- which you can do because it's a constant factor:

(c*f)' = c*f', if c is a constant.

((1/2)*x^(-1/2))' = (1/2) * (x^(-1/2))'

and then apply the power rule to x^(-1/2).
Quote:
Originally Posted by Nick View Post
Absolutely right.
You accidentally differentiated the outside and then the inside instead of differentiating the outside and then multiplying by the derivative of the inside!


I’ll finish it up the original way, since I see now what the more intelligent way looks like...

So on the step where I messed up the chain rule, it should really be:\[-1(2\sqrt{x})^{-2}*(2*\frac{1}{2}*x^{-1/2})\]Which really looks like:\[\dfrac{-1}{(2\sqrt{x})^{2}}*\dfrac{1}{\sqrt{x}}\]\[\dfrac{-1}{4x}*\dfrac{1}{\sqrt{x}}\]\[\dfrac{-1}{4x^{3/2}}\]Ok, that seems much better
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Old 2018-12-19, 22:27   #130
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I wrote down some of the practice problems from calculus, so I worked them out on paper and took a picture to make sure I’m doing them correctly.

Thanks for the help!

Click image for larger version

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ID:	19462

Last fiddled with by jvang on 2018-12-19 at 22:31 Reason: typing is hard
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Old 2018-12-20, 08:35   #131
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Well, for the physics problem you won't really need to know integrals/antiderivatives, you can easily use the kinematic formula, \(x=x_0+v_0\cdot t+\frac12\cdot a\cdot t^2\), the first two are zero because it is free fall, the third gives you the space traveled, where acceleration is g. So, yes, \(\frac{3^2\cdot 9.81}2=44.145\) meters.


(Actually, now you just got an idea where the kinematics formulas come from, if you didn't know before - plenty of guys asking on the web where the 1/2 in front of acceleration comes from ).

Last fiddled with by LaurV on 2018-12-20 at 08:42
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Old 2018-12-20, 09:11   #132
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Quote:
Originally Posted by jvang View Post
I wrote down some of the practice problems from calculus, so I worked them out on paper and took a picture to make sure I’m doing them correctly.

Thanks for the help!

Attachment 19462
1. I would put an extra step in, to convince the grader that you know what you are doing:
\[ \lim_{x\rightarrow 0}\frac{\sin(2x)}{3x}=\lim_{x\rightarrow 0}\frac{2}{3}\cdot\frac{\sin(2x)}{2x}=\ldots \]
3. I would write \( \frac{d}{dx} \sin(x^2-1)=\ldots \) instead of just \( \sin(x^2-1)=\ldots \), otherwise you risk confusing the grader!
The same goes for questions 4, 6 & 7.

4. You have made a small mistake in this one: \((2x-1)^2=4x^2-4x+1\).
7. Here, too: \(\frac{d}{dx}(5-\sin(2x))=-2\cos(2x)\).
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