Quick Question

[confused_07]

How does the derivative of \(\displaystyle 2x - (2000/sqrt(x^2)) = (2(x^3-1000))/x^2\)

 

[arthur ohlsten]

I don't think it does.
If you integrate the answer you gave you do not get the original, if I made no errors. But check this at the "integral" site on internet. find it via google


y=2x - 2000 /sqrt[x^2]
y=-2x -2000 / [x^2]^1/2
y=-2x - 2000/x
y=-2x -2000x^-1
dy/dx = -2 +2000 x^-2
dy/dx= -2 +2000/x^2

Arthur

 

[soroban]

Hello, confused_07!

Is there a typo?

How does the derivative of \(\displaystyle \L2x - \frac{2000}{\sqrt{x^2}} \;= \;\frac{2(x^3\,-\,1000)}{x^2}\;\) . . . It doesn't

We have: \(\displaystyle \L\:f(x)\;=\;2x\,-\,\frac{2000}{x}\;=\;2x\,-2000x^{-1}\)

Then: \(\displaystyle \L\:f'(x)\;=\;2\,+\,2000x^{-2} \;=\;2\,+\,\frac{2000}{x^2}\)

. . . \(\displaystyle \L f'(x) \;-\;\frac{2x^2\,+\,2000}{x^2} \;=\;\frac{2(x^2\,+\,1000)}{x^2}\)

 

[confused_07]

I apologize: there is no square root in the denominator. It should read 2x-(2000/x^2). So, how does that becoem the derivative I stated?

 

[tkhunny]

How does the derivative of \(\displaystyle 2x - (2000/sqrt(x^2)) = (2(x^3-1000))/x^2\)

This is a very odd question, even with the correction removing the square root from the denominator. Do you mean the expression to be the derivative or is one to find the derivative of the expression? It is very unclear.

In any case, if you cannot solve this one, you must have forgotten Algebra I and I would have to wonder what you are doing in Calculus. Please add the two fractions as you would any two algebraic fractions. Common denominator - that sort of thing.

 

[confused_07]

What I am asking is how did my text book get the derivative of 2x-(2000/x^2) to be [2(x^3-1000)/x^2]?

As for your second comment, yes, I have forgotten some Algebra. One, I have been out of high school for 10 years now, and two, I unfortunately am not taking classes back to back or fulltime due to the fact that I am currently in the military. With deployments and our hectic schedule, it has taken me 2yrs to complete 4 casses. That is how I arrived at being in Calculus..... and yes, asking some questions that make me look stupid. Thank you for your understanding.

 

[tkhunny]

Part of my job, that I have assigned myself, since we are all volunteers around here, is to understand the student. Sadly, most students don't really give us much information. That is why I tend to make assessments that may seem rather harsh, at times. This almost always motivates the student to provide better information. Thank you for falling into my useful trap. I think I understand your situation. It motivates me to provide some explanations that I would not provide to students who are simply lazy. There is none here trying to make anyone look stupid and we never jump to that conclusion. We'll leave that to someone else.

Reality Check: If you have forgotten enought Algebra I that you cannot add two algebraic fractions, you will have a serious disadvantage in Calculus. Frankly, I do not know how you will survive it. You must review your algebra heavily and immediately! There are only two ways to go on this - urgency or failure. Trust me.

This remains a very odd-sounding question. Something just isn't right.

First, the derivative of 2x - 2000/(x^2) is 2 + 4000/(x^3) = (2*x^3 + 4000)/(x^3) = 2*(x^3 + 2000)/(x^3). That is suspiciously close to the expression reported as the answer in the book, but I don't really see it.

Second, without reference to a derivative, 2x - 2000/(x^2) = (2*x^3 - 2000)/(x^3) = 2*(x^3 - 1000)/(x^3). This looks suspiciously close to the answer in the book, causing me to wonder if there is just a typo or something just a little off - maybe in the denominator.

Third, the ANTI-derivative of 2x - 2000/(x^2) is (x^3 + 2000)/x. Again, sort of familiar looking, but not quite there. (Don't worry if you never have heard of an anti-derivative. You'll get there eventually and I'm just guessing.)

All this causes me to wonder if we really have the right question matched up with the right answer. If you could relay the original question, in its entirety, perhaps it could be cleared up.