Post Your Problem Here (2 Viewers)

[KeypadSDM]

Simplify and then differentiate

x * x^((x - Ln[x])/Ln[x])

 

[Grey Council]

and thats two unit maths for you. hard innit?

 

[victorling]

Originally posted by KeypadSDM
Simplify and then differentiate

x * x^((x - Ln[x])/Ln[x])

solution: x*x^((x-LNx)/LNx))

=x* x^((x/LNx)-1)

=x*(x^(x/LNx)) / x

=x*(x/LNx)


Now, we can not directly differentiate x*(x/LNx), therefore

we manipulate the power this way: x/LNx

=x/(logx/loge)

=(xloge)/(logx)

= (loge^x)/(logx)

Now the power is just a number, thus allowing us to differentiate


now, the question becomes: x^((loge^x)/(logx))

differentiationof this brings:

=((loge^x)/logx)*x^((loge^x)(logx))-1

 

[overwhelming]

Prove that 1 = 2

 

[victorling]

Originally posted by overwhelming
Prove that 1 = 2

lol i am sure u know the way to prove that

 

[Jennibeans]

Originally posted by ~*HSC 4 life*~
umm what particular topic does that q fall under?

dont worry im as confused as you. i think my class may be a little behind. i recognise calculus thats a good thing right...

we're only up to stat points

 

[KeypadSDM]

Originally posted by victorling
Long and arduous anwer

You didn't fully simplify first, which makes the problem a HELL of alot harder.

 

[Mill]

Originally posted by overwhelming
Okay victorling - help us with the following:

Show that x^2/x^4+x^2+1 is less than or equal to 1/3, for all real x.

Here's an alternative solution to Keypads.


(x^2 - 1)^2 >= 0
[clearly...]

x^4 - 2x^2 + 1 >= 0

x^4 + 1 >= 2x^2

x^4 + 1 + 2x^4 + 2 >= 2x^2 + 2x^4 + 2
[adding 2x^4 + 2 to both sides]

3x^4 + 3 >= 2x^4 + 2x^2 + 2
[simplifying]

3(x^4 + 1) >= 2(x^4 + x^2 + 1)

(x^4 + 1)/(x^4 + x^2 + 1) >= 2/3

-(x^4 + 1)/(x^4 + x^2 + 1) <= -2/3
[dividing by -1]

1 - (x^4 + 1)/(x^4 + x^2 + 1) <= 1/3
[adding 1 to both sides]

(x^4 + x^2 + 1)/(x^4 + x^2 + 1) - (x^4 + 1)/(x^4 + x^2 + 1) <= 1/3

(x^2)/(x^4 + x^2 + 1) <= 1/3
[putting LHS over common denominator]

QED

 

[Grey Council]

dum de doo.

2 unit maths got a hell of a lot harder. WHY THE HELL ARE YOU DISCUSSING THIS HERE! lol

it belongs in the 3u, maybe 4u, section. chill, broz, your freaking people out.

but nice solutions. And my compliments to victorling on solving a keypad challenge. suddenly i think that victorling might actually be a good tutor.

and keypad, be a good bloke, say well done, not "you made it a hell of a lot harder for yourself".

 

[overwhelming]

why cant u prove that 1 = 2??

some people are not very good @ maths :|

 

[Grey Council]

i think they do that over at dr. math

well, they list the reason people say anyway. But its not true. (tadadadaaaa!! surprise? )

there is some error.

 

[victorling]

Originally posted by overwhelming
why cant u prove that 1 = 2??

some people are not very good @ maths :|

thank you for your comment
well, good luck for your hsc

 

[KeypadSDM]

Originally posted by GuardiaN
and keypad, be a good bloke, say well done, not "you made it a hell of a lot harder for yourself".

Dude, he didn't answer it. He got it wrong.

In fact, it's so wrong it's scary.

 

[victorling]

Originally posted by KeypadSDM
Dude, he didn't answer it. He got it wrong.

In fact, it's so wrong it's scary.

well i admit i got it wrong then
what is the right approach then?

 

[Grey Council]

LOL! so you didn't get a keypad challenge. sorry victorling.

and my apologies to you, keypad. I thought he got it right.

 

[victorling]

Originally posted by GuardiaN
LOL! so you didn't get a keypad challenge. sorry victorling.

and my apologies to you, keypad. I thought he got it right.

 

[Grey Council]

now, don't be sad lad, don't be glum chum

the fact that you even tried put you above the rest of us.

 

[victorling]

Originally posted by GuardiaN
now, don't be sad lad, don't be glum chum

the fact that you even tried put you above the rest of us.

thank you
not everyone in this forum sees it the same way as u do

 

[Mill]

The answer is e^x

Let f(x) = x * x^((x - Ln[x])/Ln[x])
= x * e^(x - Ln[x])
= x * (e^x) * e^(-Ln[x])
= x * (e^x) * 1/x
= e^x

So now it's simplified, we differentiate:

f'(x) = e^x

 

[Mill]

Actually, I found a slightly quicker way.

Let f(x) = x * x^((x - Ln[x])/Ln[x])
= x^(x/Ln[x] - 1 + 1)
= x^(x/Ln[x])
= e^x

And again,
f'(x) = e^x


They're pretty much equivalent though.