# Alt. approach to Taylor series of derivative of arcsin(x)?

1. Feb 11, 2012

Hi there,

I was hammering out the coefficients for the Taylor Series expansion of $f(x) = \frac{1}{\sqrt{1-x^2}}$, which proved to be quite unsatisfying, so decide to have a look around online for alt. approaches.

What I found (in addition to the method that uses the binomial theorem) was an old post here claiming that a trig sub could be used (here, post #5). However, the sub. - at least as written there, definitely doesn't line up:

$x = sin(u)$, which yields $f(x) = sec(u)$. However the post seems to mix up the differentials, claiming: $\frac{df}{dx} = \frac{df}{du}\cdot\frac{dx}{du}$, which is - unless I'm entirely missing something - not correct, right? We should actually have:

$\frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} = sec(u)tan(u)\cdot \frac{1}{\sqrt{1-x^2}} = sec(u)tan(u)\cdot\frac{1}{cos(u)}$ ...

I'm hoping that I've missed something here, because the method would be stellar and super insightful if it was valid, but it seems the differentials are not handled properly - can anyone confirm this?

Cheers,

Rax

2. Feb 12, 2012

Many views - any thoughts?!

Would be very appreciative of a confirmation from anyone and/or thoughts/perspectives that might lend alternative insight.

Kind regards,

Rax

3. Feb 13, 2012

### morphism

You're right.