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23 Dec 10
00:38
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03 Jul 08
Can anyone help with this question please
Differentiate x^(1/2) from first principles. [10 marks]
I got as far as:
dy lim sqrt(x + h) - sqrt(x)
-- = ----------------------
dx h=>0 h
(not very far I know!!)
but am stuck after that
cheers
Differentiate x^(1/2) from first principles. [10 marks]
I got as far as:
dy lim sqrt(x + h) - sqrt(x)
-- = ----------------------
dx h=>0 h
(not very far I know!!)
but am stuck after that
cheers
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it's not sqrt (x+h) it's f(x+h) same with the x term
define f(x)=x^1/2 therefore f(x+h)= (x+h)^1/2 seperate the terms from the brackets and then sub into to your lim equation your h term on the bottom will cancel out leaving you with just a simple equation with an h term in it, which as h=0 will go. I would do the working for you, but have a go at it, I'll help if you get stuck :) |
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Maths minds on Betfair...you are having a laugh
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aussi your approach is what I normally try however it doesn't really work for fractional powers cos you can't simplify the brackets.
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done it..
 multiplied the fraction by (sqrt(x + h) + sqrt(x)) to get h/[h(sqrt(x + h) + sqrt(x)] which leaves 1/(sqrt(x + h) + sqrt(x))... rest is trivial |
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It's the difference of two squares.
you can make your equation ((x+h)^1/2-(x^1/2))((x+h)^1/2+x^1/2)) all divided by h((x+h)^1/2+x^1/2)) perfectly ok to do that algebraicly as you are multiplying by 1. simplified down that gets you to x+h-x/h((x+h)^1/2+x^1/2) Cancelling the h gives 1/(x+h)^1/2+x^1/2 as h tends to zero we get 1/2(x^1/2) dy/dx or f'(x) whichever you prefer = 1/2x^1/2 I hope that is clear, I hate doing these things in text, you should probably write that down slowly in proper notation, some of the brackets maybe amiss. |
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ah. I misunderstood your first post. I thought you were suggesting to mmultiply out (x+h)^1/2 uing binomial. I ended up doing it the same way as you. Thanks for your help, appreciated.
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