Preliminary mathematics marathon (1 Viewer)

[hscishard]

That's not entirely correct ...

Yes this is a red herring thus why I asked it.

This 3 unit or 2unit?

This won't fail then.
y=root cos^2x
Let u=cos^2x

Dy/du x du/dx

dy/du =1/2u^-1/2
du/dx = -2sinxcosx

Therefore dy/dx = -2sinxcosx/2u^1/2
=-sinxcosx/root(cos^2x)

 

[mirakon]

or do this

cos^2 (2x)= 0.5+0.5cos4x

then differentiate

(0.5+0.5cos4x )^1/2

which should be easy. then sub x= pi/4 voila.

 

[hscishard]

Aren't you supposed to be in school?

 

[mirakon]

um aren't you supposed to be in school?

i have the day off, exam weeks

 

[hscishard]

I am at school. lol

 

[mirakon]

oh I see, free period?

 

[cutemouse]

This 3 unit or 2unit?

Probably 3U.

 

[Trebla]

Think absolute values.

 

[hscishard]

Think absolute values.

How do you differentiate absolutes?

y=|x|
y'=|1|
So uhhh it's always one?

 

[nikkifc]

y=root cos^2x
Let u=cos^2x

Dy/du x du/dx

dy/du =1/2u^-1/2
du/dx = -2sinxcosx

Therefore dy/dx = -2sinxcosx/2u^1/2
=-sinxcosx/root(cos^2x)

I think that is the correct expression for dy/dx. Now what happens if you sub x=pi/4?

 

[Trebla]

How do you differentiate absolutes?

y=|x|
y'=|1|
So uhhh it's always one?

Think about what you know about that graph. Is the slope of it always equal to 1 across the entire domain?

 

[Entrant]

Differentiating |x|
gives SIGNUM X
jeez.
OR x divided by absolute x
think about it.

 

[hscishard]

Think about what you know about that graph. Is the slope of it always equal to 1 across the entire domain?

Yea exactly, it's -1 and 1 only.

Differentiating |x|
gives SIGNUM X
jeez.
OR x divided by absolute x
think about it.

SIGNUM...what?
|x|/x...=+/- 1 wtf
I don't see your point.

 

[hscishard]

I think that is the correct expression for dy/dx. Now what happens if you sub x=pi/4?

The curve shoots up.
Gradient is like..infinity therefore undefined.

 

[hscishard]

Ok so..
y=root x^2
uhhh...
Let u = x
dy/du x du/dx
=1 x 1 = 1.

y^2 = x^2 ????
2y(dy/dx) = 2x
dy/dx = x/y
dy/dx = x/|x|

= +/-1... how are we meant to sub in the x values.
Ooo I get it. -1 when x<0 and 1 when x>0


So how do you do it...the 3 unit way

 

[Entrant]

Yea exactly, it's -1 and 1 only.


SIGNUM...what?
|x|/x...=+/- 1 wtf
I don't see your point.

duh.
abosolute x only has gradients 1 and -1
use your head.
if you differentiate |x|
you get
x/|x|
because in |x|
the gradient is either 0, 1 or -1
thats why signum was introduced.

 

[hscishard]

what is signum..
I don't think there is 0 gradient. It's a critical point at x=0.

 

[DNETTZ]

Hmm, for y=|x| at x=0 the gradient is undefined isnt it?

Because

y'(x) = sqrt(x^2)/x

Therefore cannot divide by 0?
</pre>

 

[hscishard]

Man how'd you guys get the x in the denom.

 

[DNETTZ]

Reading above posts, i think thats a bit odd the signum rule gives no indication for x=0 does it for y=|x|?