# Thread: Derivative

1. ## Derivative

Hello All,

I've calculated the following answer:

$\frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}$

And here is the answer in the back of the book:

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

Can someone please help show me how to get from A to B? Thanks

2. Hi

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

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

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

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

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

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

3. Thank You very much, I know that must have been a lot of typing!