Ironworks Gaming Forum - Maths Revision

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Quote:

Originally posted by Seraph:
Here is how I would figure out f'(x) and f''(x):
f(x) = 3^x
f(x) = e^(x * ln(3))
f'(x) = d(e^(x * ln(3)))/dx
f'(x) = e^(x * ln(3)) * ln(3)
f'(x) = 3^x * ln(3)
f''(x) = d(3^x * ln(3))/dx
f''(x) = d(e^(x * ln(3)) * ln(3))/dx
f''(x) = ln(3) * d(e^(x * ln(3)))/dx
f''(x) = ln(3) * ln(3) * 3^x

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img]

Quote:

Originally posted by Gangrell:
<blockquote>quote:<hr />Originally posted by Seraph:
Here is how I would figure out f'(x) and f''(x):
f(x) = 3^x
f(x) = e^(x * ln(3))
f'(x) = d(e^(x * ln(3)))/dx
f'(x) = e^(x * ln(3)) * ln(3)
f'(x) = 3^x * ln(3)
f''(x) = d(3^x * ln(3))/dx
f''(x) = d(e^(x * ln(3)) * ln(3))/dx
f''(x) = ln(3) * d(e^(x * ln(3)))/dx
f''(x) = ln(3) * ln(3) * 3^x

Oh man, I do math at university level, and someone mistook the inverse for the derivative! :D :D :D :D

No offense, there are people who don't know what the '2' above the x means. ;)

I don't know if anyone is interested, but my girlfriend was able to come up with a totally different way to get f'(x).

f(x) = 3^x
ln(f(x)) = ln(3^x)
ln(f(x)) = x*ln(3)
d(ln(f(x)))/dx = d(x*ln(3))/dx
d(ln(f(x))/df(x) * d(f(x))/dx = d(x*ln(3))/dx
1/f(x) * f'(x) = ln(3)
f'(x) = f(x) * ln(3)
f'(x) = 3^x * ln(3)
You should then be able to apply the same procedue to get f''(x).

Quote:

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img] [/QUOTE]I'm with you, but I don't envy them. Anything with x's are too much for my brain to handle. [img]tongue.gif[/img]

Seraph, your girlfriend got the right answer and differentiated the function correctly.

Then function is question is an exponential function; in its generic form looks like this: f(x) = k ^ x, with k being a non-negative numbers (usually). The first derivative takes the form f'(x) = (ln k) * (k ^ x). Multiple derivatives have this form: (f(n))(x) = ((ln k)^n) * (k ^ x).

If it helps, you could always think of letting k = e, because finding derivatvies for e^x is really easy. [img]graemlins/beigesmilewinkgrin.gif[/img]

Interestingly, if you let k = sqrt(2), begin with an x-value of 1, and iterate the function (the x-value of the next step is the f(x)-value from the last step) then the iterated function values converge to sqrt(2). [img]graemlins/petard.gif[/img]

Quote:

Originally posted by JrKASperov:
No offense, there are people who don't know what the '2' above the x means. ;)

2*x isn't it? :D :D :D

Quote:

Originally posted by Aragorn1:
<blockquote>quote:<hr />Originally posted by Callum:
Maths multiple choice!?

*blinks*

Wow.. Sounds... difficult.

And I'm in the year below Aragorn1... though we do call it year 12/13 here.

Does your school go from yr 7-13? [/QUOTE]Yes. :D

Quote:

Originally posted by Aragorn1:
<blockquote>quote:<hr />Originally posted by Gangrell:
<blockquote>quote:<hr />Originally posted by Seraph:
Here is how I would figure out f'(x) and f''(x):
f(x) = 3^x
f(x) = e^(x * ln(3))
f'(x) = d(e^(x * ln(3)))/dx
f'(x) = e^(x * ln(3)) * ln(3)
f'(x) = 3^x * ln(3)
f''(x) = d(3^x * ln(3))/dx
f''(x) = d(e^(x * ln(3)) * ln(3))/dx
f''(x) = ln(3) * d(e^(x * ln(3)))/dx
f''(x) = ln(3) * ln(3) * 3^x

...

o.o

...

*brain explodes* I envy people that understand that crap [img]tongue.gif[/img] [/QUOTE]No you don't.

And these days that's the 'easy' stuff for me! :D
Double angle formulae, trigometric integration AHHHH! [/QUOTE]Pfft... I know double angle formula.

And that should be easy stuff for me too... and is, now that I revised. My math exam went well... except for a right cow of a last question, involving 3d calculus and vectors and god knows what else. He said we wouldn't be having anything on planes and lines in 3d space... then gave it us anyway. [img]tongue.gif[/img]

Quote:

Originally posted by Azred:
Seraph, your girlfriend got the right answer and differentiated the function correctly.

Then function is question is an exponential function; in its generic form looks like this: f(x) = k ^ x, with k being a non-negative numbers (usually). The first derivative takes the form f'(x) = (ln k) * (k ^ x). Multiple derivatives have this form: (f(n))(x) = ((ln k)^n) * (k ^ x).

Yep... I know that know... yay for my extra three hours of revision! :D

What exam was it? Internal, external?

Internal... but my predicted grades will kinda hinge on this. [img]tongue.gif[/img]