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

In summary: So, in summary, the derivative of (sin(x* theta))/(2x-1) is [(-2sin(x*theta))/(2x-1)^2] because the quotient rule is derived from the product rule.
  • #1
avocadogirl
53
0
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.
 
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  • #2
Are you taking the derivative with respect to x or to theta?
 
  • #3
with respect to x
 
  • #4
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
avocadogirl said:
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.
 

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

1. What is the derivative of (sin(x*theta))/(2x-1)?

The derivative of (sin(x*theta))/(2x-1) is given by the quotient rule, where the numerator's derivative is (cos(x*theta))*theta and the denominator's derivative is 2. The final derivative can be simplified to [(2x-1)*(cos(x*theta))*theta - (sin(x*theta))*2]/[(2x-1)^2].

2. How do you simplify the derivative of (sin(x*theta))/(2x-1)?

To simplify the derivative of (sin(x*theta))/(2x-1), you can first distribute the (cos(x*theta))*theta term in the numerator and combine like terms. Then, using the quotient rule, you can combine the remaining terms in the numerator and denominator to get a simplified derivative.

3. What is the domain of (sin(x*theta))/(2x-1)?

The domain of (sin(x*theta))/(2x-1) is all real numbers except for x = 1/2, as this would result in a division by zero. Additionally, the domain is limited by the domain of the sine function, which is all real numbers.

4. How can the derivative of (sin(x*theta))/(2x-1) be used in real-world applications?

The derivative of (sin(x*theta))/(2x-1) can be used in various real-world applications, such as physics and engineering, to find the rate of change of a function with respect to x. It can also be used to find maximum and minimum values of a function.

5. Can the derivative of (sin(x*theta))/(2x-1) be expressed in a different form?

Yes, the derivative of (sin(x*theta))/(2x-1) can be expressed in a different form by using trigonometric identities. For example, you can rewrite the derivative as [(cos(x*theta))*theta - (sin(x*theta))*2]/[(2x-1)^2] and then use the double-angle formula for sine to get [(cos(x*theta))*theta - (sin(2x*theta))*sin(theta)]/[(2x-1)^2].

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