Maths Help Please! :)

[[TR]SweetSpotOz]

.

 

[gumbald]

42?

 

[ScarySquirrel]

I never undertood maths having words in it.

I hate maths.

 

[Welshy]

Can't remember off the top of my head, but don't you use the Quotent rule or something?

 

[Dave]

y''= [-sinx(1+sinx) - (cosx)(cosx)]/(1+sinx)^2
y'' = [-sinx - [(sinx)^2 + (cosx)^2)] / (1 + sinx)^2
= -(sinx + 1) / (1 + sinx)^2
= -1 / (1 + sinx)

Only simplification is (sinx)^2 + (cosx)^2 = 1, although I guess you figured it out by now

 

[Squark]

Can't remember off the top of my head, but don't you use the Quotent rule or something?

Sounds the right route to me

..snip..

That looks right

 

[Dave]

Sounds the right route to me

He has the right answer, it's just in the non-simplified form.

 

[Squark]

True, I was working out at the same time.

Just have to remember trig identities

 

[[TR]SweetSpotOz]

Z

 

[Majago]

d/dx[arctan(2x)/(1+4x^2)]

--> y" = [2/(1+4x^2).(1+4x^2) - 8(x)arctan(2x)] /(1+4x^2)^2

--> y" = (2 - 8(x)arctan(2x)) / (1+4x^2)^2

so yep, get the same as your edit

 

[[TR]SweetSpotOz]

Z

 

[[TR]SweetSpotOz]

.

 

[[TR]SweetSpotOz]

,

 

[titaniumx3]

Anyone?

After using the chain rule you should have:

1/[sqrt(1 - (1/sqrt(1+x))^2))] * -1/[2*(1+x)^(3/2)]

=

1/[sqrt(1-1/(1+x))] * -1/[2*(1+x)^(3/2)]

=

-1/[sqrt(x/1+x)] * 1/[2(1+x)^(3/2)]

=

-1 /[ 2*(x)^(1/2) * (1+x)^(-1/2) * (1+x)^(3/2)]

=

-1/[2*sqrt(x)*(1+x)]

which is equivalent to what stated on the site.

 

[[TR]SweetSpotOz]

.

 

[titaniumx3]

I get lost between these two lines.

OK I may have it:

1-1/(1+x) = (1+x)/(1+x) - 1/(1+x) = x/(1+x) ?

yep

 

[[TR]SweetSpotOz]

x

 

[Mansize_tissue]

I might have a crack at that problem later, markyp23. But I was just wondering what sort of Mathematics course you're doing.

 

[Fraggr]

I might have a crack at that problem later, markyp23. But I was just wondering what sort of Mathematics course you're doing.

It's either C2 or C3, looking at my books... I'm guessing C2 though.

 

[[TR]SweetSpotOz]

x