# Thread: The derivative of 1/x

1. ## The derivative of 1/x

Hi.

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

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

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

Thanks!

2. ## Re: The derivative of 1/x

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

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

$-x^{-1-1}=x^{-2}=\frac{-1}{x^2}$

3. ## Re: The derivative of 1/x

$\frac{1}{x}=x^{-1}$

$\frac{dy}{dx}(x^{-1})=-1*x^{-2}=\frac{-1}{x^2}$

4. ## Re: The derivative of 1/x

Not sure what you mean by trying to solve with conventional algebra. Using the definition of the derivative you have:

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

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

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

To make a space use "\ " - that's backslash followed by a space: "a b c" yields $a b c$, wheras "a \ b \ c" yields $a \ b \ c$.

5. ## Re: The derivative of 1/x

Thanks guys. I see how it works with this rule, but can you put it up in the fundamental formula: $\frac{f(x)\frac{1}{x+\triangle x}-f(x)}{\triangle x}$

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..

6. ## Re: The derivative of 1/x

Originally Posted by Paze
Thanks guys. I see how it works with this rule, but can you put it up in the fundamental formula: $\frac{f(x)\frac{1}{x+\triangle x}-f(x)}{\triangle x}$

Originally Posted by Paze
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.

7. ## Re: The derivative of 1/x

Originally Posted by ebaines

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!

8. ## Re: The derivative of 1/x

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}

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

Thanks.

9. ## Re: The derivative of 1/x

Originally Posted by ebaines
Not sure what you mean by trying to solve with conventional algebra. Using the definition of the derivative you have:

$\frac {d(\frac 1 x)}{dx} = \lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h = \lim _{h \to 0} \frac {x - (x+h)}{x(x+h) h} = \lim _{h \to 0} \frac {-1} {x(x+h)} = \frac {-1} {x^2}$
I mean in this..Link doesn't seem to work. So sorry for the confusion.

10. ## Re: The derivative of 1/x

When you have an expression of the form $\frac 1 a - \frac 1 b$ it may be useful to multiply the $\frac 1 a$ term by $(\frac b b )$ and the $\frac 1 b$ term by $( \frac a a )$:

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

The detailed steps are:

$\lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h$

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

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

Hope this helps.

11. ## Re: The derivative of 1/x

Originally Posted by ebaines
When you have an expression of the form $\frac 1 a - \frac 1 b$ it may be useful to multiply the $\frac 1 a$ term by $(\frac b b )$ and the $\frac 1 b$ term by $( \frac a a )$:

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

The detailed steps are:

$\lim _{h \to 0} \frac {\frac 1 {x+h} - \frac 1 x} h$

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

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

Hope this helps.
That clears it up! Thank you very much.