# Math Help - Help deriving an equation

1. ## Help deriving an equation

I need to derive this function:

$\frac{2a}{1-x^2-a^2}$

$a$ is a constant

This is how i tried to solve it:

$2a(1-x^2-a^2)^{-1}$

$\frac{4ax}{(1-x^2-a^2)^2}$

Thanks

2. Originally Posted by sebasto
I need to derive this function:

$\frac{2a}{1-x^2-a^2}$

$a$ is a constant

This is how i tried to solve it:

$2a(1-x^2-a^2)^{-1}$

$\frac{4ax}{(1-x^2-a^2)^2}$

Thanks
I think what you mean is you want to DIFFERENTIATE (not derive) that EXPRESSION (not function!)

Remember is: If $y(x) = \frac{u(x)}{v(x)}$, then $y'(x) = \frac{v(x) u'(x) - u(x) v'(x)}{(v(x))^2}$

3. Well, it was a function from the beginning, but I just omitted the f(x) and f'(x)

Here we go again then..

I need to derive this function in relation to x:

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

$a$ is a constant

This is how i tried to solve it:

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

$f'(x)=\frac{4ax}{(1-x^2-a^2)^2}$

4. Originally Posted by sebasto
Well, it was a function from the beginning, but I just omitted the f(x) and f'(x)

Here we go again then..

I need to derive this function in relation to x:

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

$a$ is a constant

This is how i tried to solve it:

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

$f'(x)=\frac{4ax}{(1-x^2-a^2)^2}$