# Differentiation .. Quadratic Rule problem :(

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

i undertsand that

$\displaystyle y=uv'+vu'$

but this question

$\displaystyle (x-9)(x-3)^{0.5}$

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

$\displaystyle (x-9)/(x-3)^{0.5} + (x-3)^{0.5}$

the real answer is $\displaystyle (3x-15)/(2(x-3)^{0.5})$
• Nov 24th 2008, 02:58 AM
mr fantastic
Quote:

Originally Posted by jimzer
hey guys i have a little problem i cant get my head around

i undertsand that

$\displaystyle y=uv'+vu'$

but this question

$\displaystyle (x-9)(x-3)^{0.5}$

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

$\displaystyle (x-9)/{\color{red}2}(x-3)^{0.5} + (x-3)^{0.5}$

the real answer is $\displaystyle 3x-15/(2(x-3)^0.5)$

The red 1/2 I added is needed. So you have $\displaystyle \frac{x-9}{{\color{red}2}(x-3)^{0.5}} + (x-3)^{0.5}$, which is the same as

$\displaystyle \frac{x-9}{2\sqrt{x-3}} + \sqrt{x-3} = \frac{x-9}{2\sqrt{x-3}} + \frac{2(x-3)}{2\sqrt{x-3}} = \frac{x-9 + 2(x-3)}{2\sqrt{x-3}} = \, ....$
• Nov 24th 2008, 03:00 AM
David24
Quote:

Originally Posted by jimzer
hey guys i have a little problem i cant get my head around

i undertsand that

$\displaystyle y=uv'+vu'$

but this question

$\displaystyle (x-9)(x-3)^{0.5}$

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

$\displaystyle (x-9)/(x-3)^{0.5} + (x-3)^{0.5}$

the real answer is $\displaystyle 3x-15/(2(x-3)^0.5)$

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 AM
jimzer
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 AM
jimzer
thanks a lot for that david :) its becoming clear
• Nov 24th 2008, 03:09 AM
mr fantastic
Quote:

Originally Posted by jimzer
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?

The derivative of $\displaystyle (x - 3)^{1/2}$ is $\displaystyle \frac{1}{2} (x - 3)^{1/2 - 1} (1) = \frac{1}{2} (x - 3)^{-1/2} = \frac{1}{2 (x - 3)^{1/2}}$.
• Nov 24th 2008, 03:10 AM
David24
Quote:

Originally Posted by jimzer
thanks a lot for that david :) its becoming clear

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 AM
jimzer
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