Help deriving an equation

**sebasto** · September 17th 2009, 02:40 PM

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:

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

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

Is that the right answer?

Thanks

**Mush** · September 17th 2009, 02:46 PM

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:

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

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

Is that the right answer?

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 $y'(x) = \frac{v(x) u'(x) - u(x) v'(x)}{(v(x))^2}$

**sebasto** · September 17th 2009, 02:50 PM

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:

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

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

Is that the right answer?

**Mush** · September 17th 2009, 02:56 PM

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:

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

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

Is that the right answer?

Again, you want to differentiate this function with respect to x.

I think you're right.

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