# fucction that I don't know how to differentiate

• Oct 15th 2009, 11:13 AM
miss.strw
fucction that I don't know how to differentiate
Differentiate the following function.
http://i37.tinypic.com/nb6tt0.gif

g ' (u) =

[LEFT]Attempt
I don't know how I got it wrong this time :(
I did
= 3u^(1/2)+2u^(1/2)
= 1.5u^(-1/2)+u(-1/2)
= (3/2)u^(-1/2)+u^(-1/2)
= 1/(3/2)u^(1/2)+1/(u^(1/2)
= 1/(sqrt(3/2)u) + 1/(sqrt(u))

I just learnt that x^-1/2 = 1/x^1/2

I don't understand what I did wrong this time :( Please help me out! thank you so much! (Shake)(Worried)
• Oct 15th 2009, 11:36 AM
earboth
Quote:

Originally Posted by miss.strw
Differentiate the following function.
http://i37.tinypic.com/nb6tt0.gif

g ' (u) =

[LEFT]Attempt
I don't know how I got it wrong this time :(
I did
= 3u^(1/2)+2u^(1/2)
= 1.5u^(-1/2)+u(-1/2)
= (3/2)u^(-1/2)+u^(-1/2)
= 1/(3/2)u^(1/2)+1/(u^(1/2)
= 1/(sqrt(3/2)u) + 1/(sqrt(u))

I just learnt that x^-1/2 = 1/x^1/2

I don't understand what I did wrong this time :( Please help me out! thank you so much! (Shake)(Worried)

Re-write the term of the function:

$\displaystyle g(u)=\dfrac1{\frac{\sqrt{3}}2 u} + \dfrac1{\sqrt{u}} = \dfrac2{\sqrt{3}} \cdot u^{-1} + u^{-\frac12}$

Then

$\displaystyle g'(u)=(-1) \cdot \dfrac2{\sqrt{3}} \cdot u^{-2} - \dfrac12 \cdot u^{-\frac32}$

That's the first rough version of the derivate. You can simplify this term, so go ahead!