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The Geniuses' Thread
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Can anyone help with this?
2 + Square Root3 is a root of the equation x^2 - 4x + c = 0, where c is a real number.
Find the value of c and write down the other root.
Ok so, I get it down to c=1, which means that the root must be 2 - Square Root3, so what I'm wondering is, if that were to come up in an exam, but the answer was c=3, how would I find the other root?
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Insect Overlord wrote: »Let c=3 and re-write the equation to suit that?
Oh I see, thanks.
It's just all to do with the minus b formula so. I don't think they'd ever put it up on ordinary level, but better to be safe than sorry. 0 -
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How odd, I was just coming here for maths help myself.
I've a note scribbled in my maths notebook, in a little faraway corner, that the exponential (e) raised to the power of a natural log, say of x (ln x) is equal to x.
i.e e^lnx = x
or, to use a constant
e^ln5 = 5
I've obviously forgotten this, as this was news to me (it's not in any of the T&T books, as far as I know). My calculator is unwilling to verify it too, helpfully letting me know "SYNTAX ERROR". So what I'm wondering is, is this true or is this some mistaken note I scribbled down eons ago?
Yup, e^lnx = x. Also works the opposite way: ln(e^x) = x.
This is a property of logs: for any constants a and x,
[latex]
\displaystyle
{log_a a^x = x}
[/latex]
and
[latex]
\displaystyle
{a^{log_a x} = x}
[/latex]
In your case, a is Euler's number. 0 -
Me again..
haha
f(x) = x - 5 over x, ie: x -5/x not x-5/x
Find F'(x), the derivative of f(x).
So I get that down to x-5x^-1
But then I was looking at the answer, and when they differentiate it, it goes to:
f'(x) = 1+5x^-2
Where does the ^-2 come from?0 -
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KnifeWRENCH wrote: »Use the quotient rule:
[latex]
\displaystyle
\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}
[/latex]
So put u = x -5 and v = x.
Then du/dx and dv/dx both equal to 1.
Now apply the quotient rule and you should end up with 5x^-2. (Unless I missed something, the 1 in the answer you quoted is a typo and shouldn't be there.)
Remember: for any number a, [latex]\frac{1}{x^a} = x^{-a}[/latex]
Btw you could also use the product rule to solve this, by rewriting it as [latex]x^{-1}(x - 5)[/latex]
Or you could just rewrite [latex]\frac{x-5}{x}[/latex] as [latex]1 - \frac{5}{x}[/latex] and differentiate it then.
Sorry I worded my original post terribly, I forgot to put the brackets in to make it seem how it should've been, so it should've been x-(5/x) as cocoa says! Thanks btw. suspect knifey's interpeted the question wrong, this is why brackets are the best!
if it's x-(5/x) then the answer given is correct and there's no need for the quotient rule, just the standard rule for differentiating powers. -1 -1 = -2, gives us -1*x^-2, but it was -1/x so we get +5*x^-2
the one is the derivative of x.
Thanks also. I'm confused though. I get how the x in x-5x^-1 turns to 1, but how do you differentiate the -5x^-1? It's the one bit that's really confusing me! 0 -
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Thanks also. I'm confused though. I get how the x in x-5x^-1 turns to 1, but how do you differentiate the -5x^-1? It's the one bit that's really confusing me!
sorry, I actually explained it really poorly =(
so the rule involved is x^a (differentiates to) -> a*x^(a-1)
in this case we have - 5*x^-1, the -5 is just a constant so can be ignored throughout the differentiation, a is -1, so a-1 = -2 so we get...
-5 * -1 * x^-2 = +5 * x^-2
hopefully that makes sense0 -
sorry, I actually explained it really poorly =(
so the rule involved is x^a (differentiates to) -> a*x^(a-1)
in this case we have - 5*x^-1, the -5 is just a constant so can be ignored throughout the differentiation, a is -1, so a-1 = -2 so we get...
-5 * -1 * x^-2 = +5 * x^-2
hopefully that makes sense
Thanks a million!
When differentiating powers is it always (a-1) yeah? 0 -
Thanks a million!
When differentiating powers is it always (a-1) yeah?
yep, that's the rule, it's in the tables on the differentiation page. I know the tables have changed, but it's still in there, it just seems all weird to me (old) =P
Basically, if you can learn to read and then apply the rules from the tables, you've got a pass at the very least
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yep, that's the rule, it's in the tables on the differentiation page. I know the tables have changed, but it's still in there, it just seems all weird to me (old) =P
Basically, if you can learn to read and then apply the rules from the tables, you've got a pass at the very least
I got an A in my mocks but I didn't do question 6 then, I'm just preparing it now in case a bad question comes up on anything else! Thanks a million for all your help. 0 -
I got an A in my mocks but I didn't do question 6 then, I'm just preparing it now in case a bad question comes up on anything else! Thanks a million for all your help.
good job! sounds like you'll do fine. When I said pass I meant at the very very least and it's 4 years since I did my leaving so I don't know stuff any more anyway
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What I meantEDIT: Also, a lot of people seem to have issues with the acronym as it seems to imply that Multiplication and Division are of a different priority (they're equally important, you just go left to right across whatever you're working out. Addition/Subtraction are similar, but of lower priority than M/D). 0
https://www.youtube.com/watch?v=u5kxybJv22I
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