1. ## Tricky derivative

I need help to do:

$\displaystyle y=\sqrt[x]{\frac{1-x}{1+x}}$

$\displaystyle \frac{1-x-x^2}{(1+x)(1-x^2)^{\frac{1}{2}}}$, but I don't think that is correct

Thanks

2. Hello, DivideBy0!

I assume we are to differentiate the fuction . . .

$\displaystyle y\:=\:\sqrt[x]{\frac{1-x}{1+x}}$
The answer says: .$\displaystyle \frac{1-x-x^2}{(1+x) (1-x^2)^{\frac{1}{2}}}$ .but I don't think that is correct
Please check the problem for typos . . . Something is terribly wrong!

If that is really an $\displaystyle x^{th}$ root, the answer cannot have a square root in it.

3. Originally Posted by DivideBy0
I need help to do:

$\displaystyle y=\sqrt[x]{\frac{1-x}{1+x}}$

$\displaystyle \frac{1-x-x^2}{(1+x)(1-x^2)^{\frac{1}{2}}}$, but I don't think that is correct

Thanks
You are correct. It isn't right.

The derivative of $\displaystyle a^x$ is $\displaystyle ln(a) \cdot a^x$, so using the chain rule on
$\displaystyle y=\sqrt[x]{\frac{1-x}{1+x}} = \left ( \frac{1 - x}{1 + x} \right ) ^{1/x}$
we obtain:
$\displaystyle y^{\prime} = ln \left ( \frac{1 - x}{1 + x} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot \left ( - \frac{1}{x^2} \right ) \cdot \left [ \frac{1 - x}{1 + x} \right ]^{\prime}$

$\displaystyle y^{\prime} = ln \left ( \frac{1 - x}{1 + x} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot \left ( - \frac{1}{x^2} \right ) \cdot \left ( \frac{-2}{(1 + x)^2} \right )$

which I would simply leave as:
$\displaystyle y^{\prime} = \left ( \frac{2}{x^2(x + 1)^2} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot ln \left ( \frac{1 - x}{1 + x} \right )$

-Dan