Maths help

SoSolid · 22 Oct 2006 at 22:43

I am confused as to what the following question is looking for. Calculate the first three derivatives of the function y = (1-3x^2)^1/2 at x =0. Now normally i would differentiate this 3 times but i am slightly confused at what it means by "at x = 0". I am not looking for someone to do the question for me, just to clarify what exactly it is I have to do.

Arcade Fire · 22 Oct 2006 at 22:46

Differentiate y with respect to x, finding y', y'' and y''', and then evaluate them at x=0.

SoSolid · 22 Oct 2006 at 22:47

That's what I was thinking I had to do, but is that not just the same thing as the Maclaurin series where u put f(0)?

SoSolid · 22 Oct 2006 at 22:59

anyone?

Arcade Fire · 22 Oct 2006 at 23:03

Yes, you'd have to do this calculation to find the coefficients of the second, third and fourth terms of the Maclaurin series for y(x) about x=0.

SoSolid · 22 Oct 2006 at 23:20

Ok I really am confused. Could someone possibly do the question so I can see the correct method?

Van_Dammesque · 22 Oct 2006 at 23:23

Try using a substitution and use the chain rule?
http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html

Then once you have them equations, put in x=0 and find to see what y is , then y' and son on.

SoSolid · 22 Oct 2006 at 23:24

Hey thanks for that, but I know how to differentiate these functions it's just that if I set x=0 then surely every evaluation will result in 0. I guess I have misinterpruted this part and would really like someone to show me how to do this.

Van_Dammesque · 22 Oct 2006 at 23:30

SoSolid said:
Hey thanks for that, but I know how to differentiate these functions it's just that if I set x=0 then surely every evaluation will result in 0. I guess I have misinterpruted this part and would really like someone to show me how to do this.

Could you write down exactly what the question is? What topic of maths are you doing? These might help.

SoSolid · 22 Oct 2006 at 23:35

its in the first post. I think i got it now, you need to use the chain and the product rule.

Arcade Fire · 22 Oct 2006 at 23:42

SoSolid said:
Hey thanks for that, but I know how to differentiate these functions it's just that if I set x=0 then surely every evaluation will result in 0. I guess I have misinterpruted this part and would really like someone to show me how to do this.

You do the differentiation first, and THEN set x=0. So you want to find y'(x), y''(x) and y'''(x), and you then set x=0 to give y'(0), y''(0) and y'''(0).

SoSolid · 23 Oct 2006 at 00:23

Yeh the problem is i am struggling to find y''. Can someone please help me to do this?

Arcade Fire · 23 Oct 2006 at 00:28

I guess you've already found y' = -3x(1-3x^2)^-1/2

You then use the product rule and the chain rule on this, so you get

y'' = -3(1-3x^2)^-1/2 - 3x(-1/2)(-6x)(1-3x^2)^-3/2

so

y'' = -3(1-3x^2)^-1/2 - 9x^2(1-3x^2)^-3/2

when you put x=0 the last term will be zero, but the first one won't.

SoSolid · 23 Oct 2006 at 01:16

Thanks a lot I get it now. What about the last one is that the same method?

Grrrrr · 23 Oct 2006 at 01:33

yep, same method...

y'' = -3(1-3x^2)^-1/2 - 9x^2(1-3x^2)^-3/2

y''' = [ (-3)(-6x)(-1/2)(1-3x^2)^-3/2 ] - [ (9x^2)(-6x)(-3/2)(1-3x^2)^-5/2 + (18x)(1-3x^2)^-3/2 ]

= -9x(1-3x^2)^-3/2 - 81x^3(1-3x^2)^-5/2 - 18x(1-3x^2)^-3/2

50p says i've made a mistake

SoSolid · 23 Oct 2006 at 01:40

where you got 54 x I think you have forgot to multiply by -3/2 as i get -81x^3.

Grrrrr · 23 Oct 2006 at 01:42

yep, you're right. I've edited it nwo