On the 5th November 1854 a smaller British and (yes) French army beat off an assault by the Russians at Inkerman in the Crimea. It was known as “The Soldier’s Battle” as men fought small engagements due to poor visibility in dense fog.
The Russians had massed 32000 men on the Allied flank and headed for the 2700 man 2nd Division, commanded today by the aggressive Pennefather. Instead of falling back in the face of superior numbers, he advanced. The British had their rifles to thank this day as they took a terrible toll on the musket armed Russian Infantry, who were hemmed in by the valley’s bottle neck shape. The British 2nd Division pushed the Russians back onto their reinforcements and should have been routed by the Russians’ numbers, but the fog and the British Light Division saved them. Three successive Russian commanders were killed in this engagement.
The Russians other 15000 men approached and assailed the Sandbag Battery, but they were routed by 300 British defenders vaulting the wall, blunting the lead Battalions, who were then attacked in the flank. More Russian attacks ensured the Battery exchanged hands several times.
The British 4th Division was not as lucky. Arriving on the field, its flanking move was itself flanked and its commander, Cathcart, killed. This enabled the Russians to advance, but not for long. They were soon driven off by French units arriving from their camps and made no more headway.
The battle was lost and they had to withdraw.
This was the last time the Russians tried to defeat the Allied troops in the field. Despite this reverse, however, the Russian attack had seriously stalled the Allies from capturing Sevastopol. They had to instead, spend one harsh winter on the heights overlooking the city, before it fell in September of 1855.
The British suffered 2573 casualties, the French 1800 and the Russians 11959.
Calculus help
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Lmao!
I’m about to start posting my homework on this site. -
Can I post my CompSci homework here?
Anyone know C++???
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d/dx(x^n) = n x^n-1 is the right rule to use.
d/dx (x^1/3) = (1/3) * x^(1/3 - 1) = (1/3) *x^(-2/3)
d/dx (x^-1/3) = (-1/3) * x^(-1/3 - 1) = (-1/3) *x^(-4/3)
You can not eliminate the 0 from the denominator
The two derivative function are both defined in all R (set of the real numbers) excluding 0, where they are not defined and 0 is a point of discontinuity.Thanks for the help.
When referring to removing 0 from the denominator, you were talking about the two derivative functions you solved for me, no?
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What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.
You said that you tried to eliminate the 0 from the denominator.
I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^(1/3) is continuous and defined in all the set of real number R. Its derivated function (1/3) * x^(-2/3) is discontinuous in the point 0, where it is not defined.
If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist. -
Can I post my CompSci homework here?
Anyone know C++???
I am an ex C/C++ programmer. After I passed to Java and nowe I am pratically an analyst/designer so I do not write code anymore! :cry:
And this is areally bad thing for me!I do not remember all the C++ syntax and rules, but you may try to ask! :-)
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What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.
You said that you tried to eliminate the 0 from the denominator.
I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.I see now, thanks.
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What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.
You said that you tried to eliminate the 0 from the denominator.
I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.I see now, thanks.
Excuse me, in the post you have quoted I wrote the derivate worng it is : (1/3)*x^(-2/3)!
But the reasoning is right.
I have corrected the original post! -
boy - i hope that Romulus gets a lot of positive Karma for this . . . .
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What is the derivative of x^(1/3) and -x(1/3)? I tried finding an equation that eliminated 0 from the denominator, but failed. I also tried using the “cheat” method (x^n ->
nx^(x-1), but couldn’t see how this was implemented.If possible could you show the algebra steps? I don’t need the calculus notation, but would like to know how it works, not just the answer.
You said that you tried to eliminate the 0 from the denominator.
I answered that it is a problem of discontinuity. Deriving a function you are worsening the function properties. Example: x^1/3 is continuous and defined in all the se of real number R. Its derivated function (1/3) * x^1/3 is discontinuous in the point 0, where it is not defined.
If you need to do some operation that uses those two derivative in the point 0, you may not use them, becasue for x = 0 the two functions are not defined, their value simply does not exist.I see now, thanks.
Excuse me, in the post you have quoted I wrote the derivate worng it is : (1/3)*x^(-2/3)!
But the reasoning is right.
I have corrected the original post!I figured that you did. I make a ton of mistakes of that ilk.
@cystic:
boy - i hope that Romulus gets a lot of positive Karma for this . . . .
And yes, he will receive positive karma.
Edit: It seems I can’t. I try and I get a message saying server verification has failed. The suggested fix does nothing.
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Prolly not a big deal. I just really, really, REALLY hate coding.
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I love coding! But in the last year I am coding less and less.
I am ususally involved in analysis and design.
I have to write a lot of documentation!I am happy that at least I am writing the docs in Latex so I still have to “compile” something!
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my brain hurts.
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@cystic:
boy - i hope that Romulus gets a lot of positive Karma for this . . . .
He should get negative Karma … how will nukchebi0 learn now?!
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@AJ:
@cystic:
boy - i hope that Romulus gets a lot of positive Karma for this . . . .
He should get negative Karma … how will nukchebi0 learn now?!
I learnt when someone showed to me how to apply derivative rules.
If you try to apply a mathematical rule and it do not work it pointless to continue to try. Mathematics is a correct science, someonw have to show you how to apply it correectly.Do, or do not. There is no try.
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There is no wrong answer in Mathematics. Just a procedure that didn’t work the way you had hoped. Sometimes a U-substitution works, sometimes Integration by Parts, sometimes FOIL, sometimes LOGs/Exponentials, etc.
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Is the thing I tried to say…
I agree with you!