# Njorl

## Guy in a red jumpsuit

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Showing Visitor Messages 1 to 1 of 1
1. Feb11-12 08:19 PM
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

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My boss didn't like me so he shot me into space.
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Gizmonics Institute
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Physicist
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"You know, Arthur, when Evil is afoot and you don't have any arms, you gotta use your head. And when Evil is ahead, and you're behind, you've gotta do the legwork! But, when you can't get a leg up, you gotta be hip! you gotta keep your chin up and kick some..." - The Tick

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