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)?

the answers say -1/(2u)(SQRTu)

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

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)?

the answers say -1/(2u)(SQRTu)

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.

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

You differentiated the function correctly:

$\displaystyle v'(u)=-\frac12 \cdot u^{-\frac32} = -\frac1{2 u^{\frac32}}= -\frac1{2 u^{1+\frac12}}=-\frac1{2 u \cdot \sqrt{u}}$

Re: Question on having the root of a variable in denominator

Ah I see, great thankyou.