Derivative of fraction

[JimmysJohnson]

G'day i understand this is beyond basic but i ran into some trouble with finding the derivative of a fraction.
From what we were taught i would always just use the formula:
(f/g)' = (f'g-fg')/g^2

So for the following equation: 1070/r i thought id just do the same thing?

However, when I use the above equation the resultant is (1070-r)/r^2
whereas in reality the answer should be -1070/r^2
I guess what im asking more is where is my understanding flawed? I saw the methodology behind getting the real answer and im not quite understanding why the (f'g-fg')/g^2 equation wouldn't bring about the same result?

Thank you for any and all help!

 

[pka]

So for the following equation: 1070/r i thought id just do the same thing?

If I faced \(f(r)=\dfrac{1070}{r}\) I would begin as \(f(r)=1070r^{-1}\)
so that \(f^{\prime}(r)=-1070r^{-2}\).

 

[Dr.Peterson]

Both ways give the same result, but only if you do them correctly.

I'd prefer to have seen your actual work, in order to find your error. But let's go through it, and I'll show you where you probably made your mistake.

You have [MATH]f(r) = 1070[/MATH] and [MATH]g(r) = r[/MATH]. Therefore [MATH]f'(r) = 0[/MATH], and [MATH]g'(r) = 1[/MATH].

So [MATH](\frac{f}{g})' = \frac{f'g-fg'}{g^2} = \frac{(0)(r)-(1070)(1)}{(r)^2} = \frac{-1070}{r^2}[/MATH].

To get [MATH]1070 - r[/MATH] in the numerator, you must have taken [MATH]f'(r) = 1[/MATH], and got signs wrong.

 

[JimmysJohnson]

Both ways give the same result, but only if you do them correctly.

I'd prefer to have seen your actual work, in order to find your error. But let's go through it, and I'll show you where you probably made your mistake.

You have [MATH]f(r) = 1070[/MATH] and [MATH]g(r) = r[/MATH]. Therefore [MATH]f'(r) = 0[/MATH], and [MATH]g'(r) = 1[/MATH].

So [MATH](\frac{f}{g})' = \frac{f'g-fg'}{g^2} = \frac{(0)(r)-(1070)(1)}{(r)^2} = \frac{-1070}{r^2}[/MATH].

To get [MATH]1070 - r[/MATH] in the numerator, you must have taken [MATH]f'(r) = 1[/MATH], and got signs wrong.

Ah thank you i see where i made my mistake, it was as exactly as you said, was treating f'(r) as 0 rather than 1. Thank you very much everyone!

 

[HallsofIvy]

No, your mistake was treating f'(r) as 1 rather than as 0!

 

[Steven G]

But doesn't 0! = 1?? I'm so confused.

 

[CurveGuy]

Jomo meant zero ... '0' ... not zero factorial.

 

[HallsofIvy]

I think you have the name wrong. I referred to the number 0, ending the sentence with an exclamation point. Jomo then, jokingly I am sure, interpreted it as 0 factorial.