# Question on having the root of a variable in denominator

• Jan 29th 2013, 01:51 AM
camjerlams
Question on having the root of a variable in denominator
Hi, Ive just been learning how to do some derivatives and have become stuck on a question because i think I dont understand some algebra.

ok question is finding the derivative

v=1/SQRT(u)
d/du 1/SQRT(u)?

I know that 1/SQRT(u) = SQRT(u)^-1 but to get u on its own would it then become u^-1.5? or u^-0.5? And I'm wondering is the answer a typo? I thought I knew the power rule and cant see how that answer can e right. I know this is calc now but Im posting here as Im missing some simple algebra/arithmetic
• Jan 29th 2013, 04:29 AM
earboth
Re: Question on having the root of a variable in denominator
Quote:

Originally Posted by camjerlams
Hi, Ive just been learning how to do some derivatives and have become stuck on a question because i think I dont understand some algebra.

ok question is finding the derivative

v=1/SQRT(u)
d/du 1/SQRT(u)?

I know that 1/SQRT(u) = SQRT(u)^-1 but to get u on its own would it then become u^-1.5? or u^-0.5? And I'm wondering is the answer a typo? I thought I knew the power rule and cant see how that answer can e right. I know this is calc now but Im posting here as Im missing some simple algebra/arithmetic

All your considerations and calcuations are correct.

$v(u)=\frac1{\sqrt{u}}=u^{-\frac12}$

You differentiated the function correctly:

$v'(u)=-\frac12 \cdot u^{-\frac32} = -\frac1{2 u^{\frac32}}= -\frac1{2 u^{1+\frac12}}=-\frac1{2 u \cdot \sqrt{u}}$
• Jan 29th 2013, 06:28 PM
camjerlams
Re: Question on having the root of a variable in denominator
Ah I see, great thankyou.