The derivative of 1/x

Nov 2012
379
19
Iceland
Hi.

I'm having problems understanding \(\displaystyle \frac{dy}{dx}\(\frac{1}{x})\)

I always get 0 when I try to solve it with conventional algebra but the answer is \(\displaystyle \frac{-1}{x^2}\)

Bonus question: How do I get the enclosing brackets in my tex code to cover \(\displaystyle \frac{1}{x}\) completely, instead of being small? Also, how do I do a space between \(\displaystyle \frac{dy}{dx}\) and \(\displaystyle \frac{1}{x}\) ?

Thanks!
 

Plato

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I'm having problems understanding \(\displaystyle \frac{dy}{dx}\(\frac{1}{x})\)
I always get 0 when I try to solve it with conventional algebra but the answer is \(\displaystyle \frac{-1}{x^2}\)
Bonus question: How do I get the enclosing brackets in my tex code to cover \(\displaystyle \frac{1}{x}\) completely, instead of being small? Also, how do I do a space between \(\displaystyle \frac{dy}{dx}\) and \(\displaystyle \frac{1}{x}\) ?
[noparse]\(\displaystyle \frac{dy}{dx}\left(\frac{1}{x}\right)\)[/noparse] gives \(\displaystyle \frac{dy}{dx}\left(\frac{1}{x}\right)\)

\(\displaystyle \frac{dy}{dx}\left(\frac{1}{x}\right)=\frac{dy}{dx}(x^{-1})\)

\(\displaystyle -x^{-1-1}=x^{-2}=\frac{-1}{x^2}\)
 
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Nov 2011
33
1
\(\displaystyle \frac{1}{x}=x^{-1}\)

\(\displaystyle \frac{dy}{dx}(x^{-1})=-1*x^{-2}=\frac{-1}{x^2}\)
 
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Jun 2008
1,389
513
Illinois
Not sure what you mean by trying to solve with conventional algebra. Using the definition of the derivative you have:

\(\displaystyle \frac {dy}{dx} = \lim_{h \to 0} \frac {f(x+h)-f(x)} h\)

\(\displaystyle \frac {d(\frac 1 x)}{dx} = \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2} \)

Bonus questions: to make the parentheses larger you can use "\left(" and "\right)", like this:
\(\displaystyle \left( \frac {-1} {x^2} \right) \)

To make a space use "\ " - that's backslash followed by a space: "a b c" yields \(\displaystyle a b c\), wheras "a \ b \ c" yields \(\displaystyle a \ b \ c\).
 
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Nov 2012
379
19
Iceland
Thanks guys. I see how it works with this rule, but can you put it up in the fundamental formula: \(\displaystyle \frac{f(x)\frac{1}{x+\triangle x}-f(x)}{\triangle x}\)

Okay I have no idea how this tex code is gonna come out seeing as I can't preview my answer, but I am asking to see the proof in the fundamental formula: (f(x+delta X) - f(x))/delta x

Edit: holy sh** that came out almost right..
 
Jun 2008
1,389
513
Illinois
Thanks guys. I see how it works with this rule, but can you put it up in the fundamental formula: \(\displaystyle \frac{f(x)\frac{1}{x+\triangle x}-f(x)}{\triangle x}\)
Did you see my reply in post #4 of this thread?

Okay I have no idea how this tex code is gonna come out seeing as I can't preview my answer, but I am asking to see the proof in the fundamental formula: (f(x+delta X) - f(x))/delta x
You should have a "Go Advanced" button to the lower right, below the box where you type your reply - click it and you'll see a preview of your post.
 
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Nov 2012
379
19
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Did you see my reply in post #4 of this thread?



You should have a "Go Advanced" button to the lower right, below the box where you type your reply - click it and you'll see a preview of your post.
I noticed it now!
Thank you and thank you all for your help!
 
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Nov 2012
379
19
Iceland
Hold on...I'm trying to get this to work.. I can't seem to get it to work!

In this: http://latex.codecogs.com/png.latex?%20\frac%20{d(\frac%201%20x)}{dx}%20=%20\lim%20_{h%20\to%200}%20\frac%20{\frac%201%20{x+h}%20-%20\frac%201%20x}%20h%20=%20\lim%20_{h%20\to%200}%20\frac%20{x%20-%20(x+h)}{x(x+h)%20h}%20=%20\lim%20_{h%20\to%200}%20\frac%20{-1}%20{x(x+h)}%20=%20\frac%20{-1}%20{x^2}

(sorry for long link)

How does the swap between numerators happen and how does x-x-h become -1? -h becomes 0??

Thanks.
 
Nov 2012
379
19
Iceland
Not sure what you mean by trying to solve with conventional algebra. Using the definition of the derivative you have:


\(\displaystyle \frac {d(\frac 1 x)}{dx} = \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2} \)
I mean in this..Link doesn't seem to work. So sorry for the confusion.
 
Jun 2008
1,389
513
Illinois
When you have an expression of the form \(\displaystyle \frac 1 a - \frac 1 b\) it may be useful to multiply the \(\displaystyle \frac 1 a \) term by \(\displaystyle (\frac b b )\) and the \(\displaystyle \frac 1 b \) term by \(\displaystyle ( \frac a a ) \):

\(\displaystyle \frac 1 a - \frac 1 b = \frac {b}{b} ( \frac 1 a) - \frac a a (\frac 1 b) = \frac {b-a}{ab}\)

The detailed steps are:

\(\displaystyle \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h \)

\(\displaystyle = \lim _{h \to 0} \frac {\frac {x}{x(x+h)} - \frac {x+h}{x(x+h)}}h\)

\(\displaystyle = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-h}{x(x+h)h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2} \)

Hope this helps.
 
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