How to Solve for the Nth Derivative of a Trigonometric Function?

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In summary, the question asks for the n'th derivative of sin(2x) and suggests that it can be found by solving 3y(n) - y'' - y = 0. Unfortunately, this doesn't work because 1/√sin can't be an answer.
  • #1
JanClaesen
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For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!
 
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  • #2


Hello,

For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!

What are you trying to do here? Are you trying to find the derivative of

sin(2x)^(-0.5) ?

What is a'd derivative?

Thanks
Matt
 
  • #3
Hi JanClaesen! :smile:

(have a pi: π and a square-root: √ :wink:)
JanClaesen said:
For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!

(we say the nth derivative, not the n'd derivative :wink:)

Do you mean "what is the nth derivative of 1/√(sin2x)?"

And what is f? :confused:
 
  • #4


Hello :smile:, I'm sorry if I was a little unclear:

So there is a certain relation for the function sin(2x)^(-1/2): the second derivative of this function plus the function itself gives the (n'th derivative)*3, the question is to determine n.
 
  • #5


Rephrase:

If [itex]y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}[/itex], solve for n. (Note [itex]y^{(n)} = \frac{d^n y}{{dx}^n}[/itex].)

I have hammered out the first 10 derivatives of y but none seem to be equal to (y'' + y)/3. The exponent of the denominator factor is of the order 1/2 + n, while the numerator is a trigonometric polynomial in cosine of order n. I do not forsee this collapsing into something nice. This exercise seems fishy.

N.B.: The expression on the left of the differential equation is

[tex]\frac{3}{(sin(2x))^{5/2}}.[/tex]​

--Elucidus
 
  • #6
Elucidus said:
If [itex]y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}[/itex], solve for n. (Note [itex]y^{(n)} = \frac{d^n y}{{dx}^n}[/itex].)

Hi JanClaesen! Hi Elucidus! :smile:

I don't think there can be a solution …

just try to solve 3y(n) - y'' - y = 0 the usual way …

that gives you a characteristic equation, of which 1/√sin can't be an answer. :redface:
 
  • #7


I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg
It's exercice 17a, the answer should be 5, perhaps alpha is a power?
 
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  • #8
JanClaesen said:
I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg
It's exercice 17a, the answer should be 5, perhaps alpha is a power?

D'oh!

Yes of course α is a power … Question 17a says so :rolleyes: …

f'' + f = 3fα

Why did you write D2(f) + f = 3*D(a) in your first post??

ok, since f = 1/√(sin(2x)), what is f'' + f?
 
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  • #9


Because I only thought of that while I was writing the newest post :smile:
I'm sorry, but I never saw a power-variable being named alpha before, that's why I thought it was perhaps a way to note the n'th derivative, which seemed to me like a quite hard thing to solve.
Hope I didn't waste too much of your time!
 
  • #10


Just curious ... what language is that question written in ... and are you from the Netherlands ?
 
  • #11


Hah nice, you're right, it's Dutch :wink:, but I'm not from the Netherlands but from its little brother, Belgium. Most people here are American I guess?
 

1. What is the domain of the function sin(2x)^(-1/2)?

The domain of this function is all real numbers except for values that make the expression inside the square root equal to zero, which results in an undefined value. In this case, the domain is all real numbers except for values that make 2x equal to odd multiples of π/2.

2. What is the range of the function sin(2x)^(-1/2)?

The range of this function is all real numbers greater than or equal to -1 and less than or equal to 1, since the sine function has a range of -1 to 1 and the reciprocal function will flip the values. However, some values may be excluded due to the domain restrictions mentioned in the previous question.

3. What is the period of the function sin(2x)^(-1/2)?

The period of this function is π, since the 2x term inside the sine function will cause it to repeat every π radians. The square root does not affect the period of the function.

4. How does the graph of sin(2x)^(-1/2) compare to the graph of sin(2x)?

The graph of sin(2x)^(-1/2) will have the same basic shape as the graph of sin(2x), but it will be flipped and stretched vertically due to the reciprocal function. The zeros of the sine function will become vertical asymptotes in the graph of sin(2x)^(-1/2).

5. Can the function sin(2x)^(-1/2) be simplified?

Yes, the function can be simplified by using the trigonometric identity cos^2(x) + sin^2(x) = 1. By substituting 2x for x, we get cos^2(2x) + sin^2(2x) = 1. Then, we can replace cos^2(2x) with 1 - sin^2(2x), giving us 1 - sin^2(2x) + sin^2(2x) = 1. This simplifies to 1 = 1, which is always true. Therefore, the simplified form of sin(2x)^(-1/2) is just 1.

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