# Need help finding the derivative.

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• Jun 21st 2009, 09:37 PM
Darkhrse99
I need to find the derivative of

$x-2 \sqrt{x-1}$

I knew to use the power rule, but I was coming up with
$1/2(x-1)^{-1/2}$ Is this the correct answer?

Thanks
Jason
• Jun 21st 2009, 09:50 PM
Chris L T521
Quote:

Originally Posted by Darkhrse99
I knew to use the power rule, but I was coming up with
$1/2(x-1)^-1/2$ Is this the correct answer?

The derivative (by power rule & chain rule) is $1-\left(x-1\right)^{-1/2}=1-\frac{1}{\sqrt{x-1}}$, where the derivative of $\sqrt{x-1}$ is $\frac{1}{2}\left(x-1\right)^{-1/2}=\frac{1}{2\sqrt{x-1}}$
• Jun 22nd 2009, 09:57 PM
Darkhrse99
Quote:

Originally Posted by Chris L T521
The derivative (by power rule & chain rule) is $1-\left(x-1\right)^{-1/2}=1-\frac{1}{\sqrt{x-1}}$, where the derivative of $\sqrt{x-1}$ is $\frac{1}{2}\left(x-1\right)^{-1/2}=\frac{1}{2\sqrt{x-1}}$

How did you know to use the chain rule?
• Jun 23rd 2009, 01:15 PM
Krizalid
chain rule is present at any moment.

for example if you want to differentiate a simple function as $x^2,$ then that is $2x\cdot(x)'=2x,$ but that it's quite simple and then irrelevant to write, but having $\sqrt{x-1}$ we can think this as the composition by letting $f(x)=\sqrt x$ and $g(x)=x-1,$ so by differentiating $f\big(g(x)\big)$ we get $f'\big(g(x)\big)\cdot g'(x).$ According to this we have that $\sqrt{x-1}$ can be expressed as $h(x)=f\big(g(x)\big)=\sqrt{x-1},$ and then by differentiating we get (following up on what I said before) $h'(x)=\frac1{2\sqrt{x-1}}\cdot(x-1)'=\frac1{2\sqrt{x-1}}.$