derivative method dy/dx

crypt0s · #1 $Old$ September 29th, 2009, 10:59 PM

After much searching on google and no help, I decided to come here:

I'm searching for why/how the derivative of 1/t is equal to -(1/t^2)

showing work would probably instantly clear this up for me. I work with computers and code so having the rules as to Why this is that will help me more than anything.

Abstract explanations abound on google. If you could explain using the example I'd be forever grateful.

dogmom09 · #2 $Old$ September 29th, 2009, 11:05 PM

1/t is equal to -(1/t^2)

1/t is the same as 1t^-1
If the t is on the bottom it has a negative exponent.
Take the derivative of that by multiplying the 1 in front of the t by the -1 in the exponent. This will give you -1t^-2 or -1/t^2

Did that clear it up at all?

Bruno J. · #3 $Old$ September 29th, 2009, 11:07 PM

From the definition of the derivative :

$\lim_{h\rightarrow 0}\frac{1/(t+h)-1/t}{h} = \lim_{h\rightarrow 0}\frac{(t-(t+h))/(t(t+h))}{h}= \lim_{h\rightarrow 0}\frac{-h}{t(t+h)h}$

$= \lim_{h\rightarrow 0}\frac{-1}{t(t+h)} = -1/t^2$

Bruno J. · #4 $Old$ September 29th, 2009, 11:07 PM

Quote:

Originally Posted by dogmom09 $View Post$

1/t is equal to -(1/t^2)

Huh? The rest is fine but this is obviously wrong.

crypt0s · #5 $Old$ September 29th, 2009, 11:48 PM

Quote:

Originally Posted by dogmom09 $View Post$

1/t is equal to -(1/t^2)

1/t is the same as 1t^-1
If the t is on the bottom it has a negative exponent.
Take the derivative of that by multiplying the 1 in front of the t by the -1 in the exponent. This will give you -1t^-2 or -1/t^2

Did that clear it up at all?

im more confused. i thought that dy/dx was a kind of shortcut?

mr fantastic · #6 $Old$ September 30th, 2009, 12:17 AM

Quote:

Originally Posted by crypt0s $View Post$

im more confused. i thought that dy/dx was a kind of shortcut?

Read post #3. It gave you the reason.

crypt0s · #7 $Old$ October 1st, 2009, 09:25 PM

i figured it out. Its just another way to say: derivative of.

Bruno J. · #8 $Old$ October 1st, 2009, 10:04 PM

Yes and no; almost everybody, in their introductory calculus course, learns that $\frac{dy}{dx}$ is just another notation for $f'(x)$ ; but soon enough you will start seeing things like $dy=dx$ , which you would not expect to have meaning if $\frac{dy}{dx}$ was just a notation.

In fact, very few people seem to know what $dx$ and $dy$ actually mean. Most people just treat them like regular variables with a kind of algebraic faith. Of course they can be defined as such - just variables - but then $\frac{dy}{dx}$ ceases to be the derivative. Most people get around this mess by saying that they are "infinitesimal" variables - variables which can only take infinitely small values. This is more correct but still quite sloppy.

I'd go as far as to say that what Leibniz's notation really means is almost taboo. It took me quite a while to be comfortable with differentials, and the only way in which I can rigorously define them is quite complicated.

Matt Westwood · #9 $Old$ October 1st, 2009, 11:24 PM

Quote:

Originally Posted by Bruno J. $View Post$

Yes and no; almost everybody, in their introductory calculus course, learns that $\frac{dy}{dx}$ is just another notation for $f'(x)$ ; but soon enough you will start seeing things like $dy=dx$ , which you would not expect to have meaning if $\frac{dy}{dx}$ was just a notation.

In fact, very few people seem to know what $dx$ and $dy$ actually mean. Most people just treat them like regular variables with a kind of algebraic faith. Of course they can be defined as such - just variables - but then $\frac{dy}{dx}$ ceases to be the derivative. Most people get around this mess by saying that they are "infinitesimal" variables - variables which can only take infinitely small values. This is more correct but still quite sloppy.

I'd go as far as to say that what Leibniz's notation really means is almost taboo. It took me quite a while to be comfortable with differentials, and the only way in which I can rigorously define them is quite complicated.

Not sure I agree. Once you learn what the notation means, and learn what the rules are for the manipulation of the various symbols, then it's a useful notation. The important thing here is *learn what the rules are*, like any other formal system. A lot of the time you just want it to work, you're not that interested in the "why" of it, you're using a tool to get a result.