hey guys i have a little problem i cant get my head around

i undertsand that

but this question

thats (x-3)^0.5 which is root (i dont know how to root in the maths tags

but the answer i keep getting is

the real answer is

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- Nov 24th 2008, 02:27 AMjimzerDifferentiation .. Quadratic Rule problem :(
hey guys i have a little problem i cant get my head around

i undertsand that

but this question

thats (x-3)^0.5 which is root (i dont know how to root in the maths tags

but the answer i keep getting is

the real answer is - Nov 24th 2008, 02:58 AMmr fantastic
- Nov 24th 2008, 03:00 AMDavid24
hey mate,

as you stated you have an equation of the form

y = u(x)v(x) for which you want to find the derivative which as you stated is solved using the product rule,

i.e.,

y'(x) =u'(x)v(x) + u(x)v'(x)

here,

(i) u(x) = (x-9) --> u'(x) = 1

(ii) v(x) = (x-3)^(1/2) --> v'(x) = (1/2)*(x-3)^(-1/2) (I'm assuming you know the chain rule - please advise me if you dont and I'll be more than happy to explain)

Therefore

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

= (1/2)*(3x - 15)/(x-3)^(1/2)

Hope this helps,

Let me know if you require any further assistance,

David - Nov 24th 2008, 03:00 AMjimzer
thanks so much again youve helped me so much ..

props to you if i get a sweet mark on the exam :)

but 1 question. why is the 2 needed? - Nov 24th 2008, 03:02 AMjimzer
thanks a lot for that david :) its becoming clear

- Nov 24th 2008, 03:09 AMmr fantastic
- Nov 24th 2008, 03:10 AMDavid24
hey mate,

not a problem - just in reference to the (1/2) component I believe you were having troubles with, as you would know it composed in the derivative of,

v(x) = (x-3)^(1/2)

Here we employ the Chain rule, i.e. let w = x-3 then

v = w^(1/2) --> v'(w) = (1/2)w^(-1/2) (i.e. d/dx (x^n) = nx^(n-1) )

and w'(x) = 1

Therefore by the Chain Rule,

v'(x) = v'(w)*w'(x) or v'(x) = (dv/dw)*(dw/dx)

Thus,

v'(x) = (1/2)w^(-1/2) * (1)

Recall w = x-3

Therefore

v'(x) = (1/2)(x-3)^(-1/2)

Hope this helps,

Regards,

David

ps - are you good with your logarithms? there is another technique which makes alot of these nasty derivates simpler to handle. - Nov 24th 2008, 03:12 AMjimzer
thanks a lot mr fantastic and david, you guys have been of so much help

and i'm okay with my logs but i rather not delve into them right this moment! might get a headache

take care guys, thanks so much for all the help, it is greatly appreciated