# Square root derivative

• Nov 18th 2009, 06:57 PM
Bacterius
Square root derivative
Hi, I'm trying to derivate this function : $\displaystyle f(x) = \sqrt{2x - 1}$. I'm applying this principle : if $\displaystyle f = a^n$ then $\displaystyle f' = na^{n - 1}$. So there I go :

$\displaystyle f(x) = \sqrt{2x - 1}$

$\displaystyle f(x) = (2x - 1)^{\frac{1}{2}}$

Now I derivate :

$\displaystyle f'(x) = \frac{1}{2} (2x - 1)^{\frac{-1}{2}}$

$\displaystyle f'(x) = \frac{1}{2} \frac{1}{(2x - 1)^{\frac{1}{2}}}$

$\displaystyle f'(x) = \frac{1}{2} \frac{1}{\sqrt{2x - 1}}$

$\displaystyle f'(x) = \frac{1}{2 \sqrt{2x - 1}}$

But the book says that's not right and the answer should be :

$\displaystyle f'(x) = \frac{2}{2 \sqrt{2x - 1}} = \frac{1}{\sqrt{2x - 1}}$

Does anyone see when I went wrong ? Thanks a lot ...
• Nov 18th 2009, 11:15 PM
earboth
Quote:

Originally Posted by Bacterius
Hi, I'm trying to derivate this function : $\displaystyle f(x) = \sqrt{2x - 1}$. I'm applying this principle : if $\displaystyle f = a^n$ then $\displaystyle f' = na^{n - 1}$. So there I go :

$\displaystyle f(x) = \sqrt{2x - 1}$

$\displaystyle f(x) = (2x - 1)^{\frac{1}{2}}$

Now I derivate :

$\displaystyle f'(x) = \frac{1}{2} (2x - 1)^{\frac{-1}{2}}$ <<<<<<< here! You forgot to apply the chain rule

$\displaystyle f'(x) = \frac{1}{2} \frac{1}{(2x - 1)^{\frac{1}{2}}}$

$\displaystyle f'(x) = \frac{1}{2} \frac{1}{\sqrt{2x - 1}}$

$\displaystyle f'(x) = \frac{1}{2 \sqrt{2x - 1}}$

But the book says that's not right and the answer should be :

$\displaystyle f'(x) = \frac{2}{2 \sqrt{2x - 1}} = \frac{1}{\sqrt{2x - 1}}$

Does anyone see when I went wrong ? Thanks a lot ...

$\displaystyle f(x) = (2x - 1)^{\frac{1}{2}}$

$\displaystyle f'(x) = \underbrace{\frac{1}{2} (2x - 1)^{\frac{-1}{2}}}_{\text{der. of the square-root}} \cdot \underbrace{2}_{\text{der. of the radicand}}$
• Nov 23rd 2009, 08:42 PM
Bacterius
Thanks a lot ! (Happy)