Njorl

Hi Njorl -

I saw an old post you made some time ago & was curious as to it's accuracy. The thread is (http://www.physicsforums.com/showthread.php?t=7895) and addresses Taylor series for the derivative of arcsin(x).

In the post, you wrote the following:

"to get f'(x)=dy/dx for the taylor series, first calculate dy/du then multiply by dx/du

dy/du=sec(u)tan(u)

multiplying the left by dx/du and the right by cos(u) (remember, dx/du=cos(u) ) yields,

f'(x)=dy/dx=tan(u)"

I don't follow this development as it is written, because (dy/du)*(dx/du) /= dy/dx. (I don't know the common symbol for 'not equal to', so I've used "/="). Don't we actually need dy/dx = dy/du * du/dx ? Where du/dx = d(arcsin(x))/du = 1/sqrt(1-x^2) = 1/cos(u).

I've tried this several times over, but instead of having the nice cancellation of cos(u) terms that you claim, we seem to end up with a [sec(u)]^2 * tan(u). Could you clarify, please - I've very curious!


Cheers,


Rax