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

  • #1
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))
_______________________________

(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.
 

Answers and Replies

  • #2
534
1
Are you taking the derivative with respect to x or to theta?
 
  • #4
534
1
My TI-89 gives me
[tex]\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}.[/tex]
 
  • #5
okay...I see why I was getting that answer from the calculator. If you wanted to do the differentiation by hand, would you use the product rule to derive the terms with which you would use the quotient rule? Or, can you even do it that way?
 
  • #6
djeitnstine
Gold Member
614
0
okay...I see why I was getting that answer from the calculator. If you wanted to do the differentiation by hand, would you use the product rule to derive the terms with which you would use the quotient rule? Or, can you even do it that way?
You can do it either way. Quotient rule is simply a derivative of the product rule.
 

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