Square root derivative

**Bacterius** · November 18th 2009, 06:57 PM

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

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

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

Now I derivate :

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

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

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

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

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

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

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

**earboth** · November 18th 2009, 11:15 PM

Originally Posted by Bacterius

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

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

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

Now I derivate :

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

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

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

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

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

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

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

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

$f'(x) = \underbrace{\frac{1}{2} (2x - 1)^{\frac{-1}{2}}}_{\text{der. of the square-root}} \cdot \underbrace{2}_{\text{der. of the radicand}}$

**Bacterius** · November 23rd 2009, 08:42 PM

Thanks a lot !

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