# Differentiating (sin(x*theta))/(2x-1)?

1. Feb 8, 2009

Why does the derivative of (sin(x* theta))/(2x-1) equal [(-2sin(x*theta))/(2x-1)^2]?

I'm an undergraduate physics student with a shaky math background, who's enrolled in an advanced Physics math class for which I lack prerequisites. I'm working an infinite series where the general term comes to (sin(x*theta))/(2x-1) and, applying the quotient rule to take the derivative, I get the result:

(2x-1)(cos(x*theta))-(2sin(x*theta))
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(2x-1) ^2

But, my TI-89 eliminates the first part of the numerator, giving me
[(-2sin(x*theta))/(2x-1)^2]

I don't really need help with the problem I'm solving, I would just like to know why the cos term is eliminated. I can't seem to find an identity to explain this, or, at least, I can't see the correlation. Any assistance is sincerely appreciated.

Thank you.

2. Feb 8, 2009

Are you taking the derivative with respect to x or to theta?

3. Feb 8, 2009

with respect to x

4. Feb 8, 2009

My TI-89 gives me
$$\frac{d}{dx} \left( \frac{\sin(x \theta)}{2x - 1} \right) = \frac{\theta \cos(\theta x)}{2x - 1} - \frac{2 \sin(\theta x)}{(2x - 1)^2}.$$

5. Feb 8, 2009