1. ## Derivative

How do I get the correct derivative for this function?

$y = \frac{2x-1}{(\sqrt{x+1})}$

I got the first part of $\frac{2}{\sqrt{x+1}}$ but cant seemed to get the second part of it using the quotient rule $\frac{2x-1}{2(x+1)^{3/2}}$

y' in the answer's is supposedly

$y ' = \frac{2}{\sqrt{x+1}}-\frac{2x-1}{2(x+1)^{3/2}}$

Can someone please show me how to get to that using the quotient rule?

2. Originally Posted by dwat
How do I get the correct derivative for this function?

$y = \frac{2x-1}{(\sqrt{x+1})}$

I got the first part of $\frac{2}{\sqrt{x+1}}$ but cant seemed to get the second part of it using the quotient rule $\frac{2x-1}{2(x+1)^{3/2}}$

y' in the answer's is supposedly

$y ' = \frac{2}{\sqrt{x+1}}-\frac{2x-1}{2(x+1)^{3/2}}$

Can someone please show me how to get to that using the quotient rule?
$y = \frac{2x - 1}{\sqrt{x + 1}} = \frac{2x - 1}{(x + 1)^{\frac{1}{2}}}$.

Using the Quotient rule:

$\frac{dy}{dx} = \frac{(x + 1)^{\frac{1}{2}}\frac{d}{dx}(2x - 1) - (2x - 1)\frac{d}{dx}\left((x + 1)^{\frac{1}{2}}\right)}{\left((x + 1)^{\frac{1}{2}}\right)^2}$

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

You could try to simplify that or write it with radicals if you wish...