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Homework Help: Chain rule problem

  1. Sep 17, 2011 #1
    I always get muddled when I'm dealing with chain rule of any degree of complexity and also when dealing with powers of trig. functions - this problem contains both:

    find [tex]\frac{\partial n}{\partial A}[/tex] and [tex]\frac{\partial n}{\partial D}[/tex] of the following function:


    also find [tex]|\frac{\partial n}{\partial D}|^{2}[/tex] [tex]|\frac{\partial n}{\partial A}|^{2}[/tex]

    My Attempt:

    [tex]\implies n=\sin(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}[/tex]
    [tex]\implies\frac{\partial n}{\partial A}=\frac{1}{2}\cos(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}+\sin(\frac{A+D}{2})(-1)[\sin(\frac{A}{2})]^{-2}(\frac{1}{2})\cos(\frac{A}{2})[/tex]



    [tex]\implies|\frac{\partial n}{\partial A}|^{2}=\frac{1}{4\sin^{2}(\frac{A}{2})}[\cos^{2}(\frac{A+D}{2})+\sin^{2}(\frac{A+D}{2})\arctan^{2}(\frac{A}{2})-2\cos(\frac{A+D}{2})\sin(\frac{A+D}{2})\arctan(\frac{A}{2})][/tex]

    *Correction that last arctan term should read [tex]\arctan(\frac{A}{2})[/tex]

    At this point I tried to simplify the trig. a bit.... (but a don't think I did a great job!):


    Main issues I have with the above are: I am never any good with chain rule and also re. the trig. I keep getting confused as to whether for example the function:
    is in all cases completely equivalent to:

    ie. are there some cases where they are not equivalent? - this confusion may or may not have caused errors in my attempt above...

    OK, now [tex]\frac{\partial n}{\partial D}[/tex] is a little easier:

    [tex]\frac{\partial n}{\partial D}=\frac{1}{2}\cos(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}[/tex]

    [tex]\implies|\frac{\partial n}{\partial D}|^{2}=\frac{1}{4}\cos^{2}(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-2}[/tex]

    All advice and corrections are greatly appreciated.
    Last edited: Sep 18, 2011
  2. jcsd
  3. Sep 17, 2011 #2
    Your original function is pretty complex.


    I will show you how to find find

    [tex]\frac{\partial n}{\partial A}[/tex] of [tex]n=sin(\frac{A+D}{2})[/tex]

    after seeing this then you can apply it to the problem as a whole using quotient rule.

    [tex]\frac{\partial n}{\partial A}[/tex]=[cos(0.5A+0.5D)](0.5)

    Think, this is a partial derivative with respect to A, this means that A is the only variable and that everything else including D will be held constant.

    Chain rule goes like this: derivative of the outside, inside stays the same then derivative of the inside, repeat until you don't have anymore insides.

    The outside was sin() and the inside was (0.5A+0.5D) and the derivative of the inside as (0.5)
  4. Sep 17, 2011 #3


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    I would apply the sum rule, and write n in terms of separate trigonometric functions of A/2 and D/2.

    And sin^n(A) means [sin(A)]^n.

  5. Sep 18, 2011 #4
    ... so going by the above comments, I think I used the chain rule correctly (@ehilld I find chain rule easier than sum rule mostly!) - can anyone suggest tips on if there is a way I can simplify the trig. a little?
  6. Sep 18, 2011 #5


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    Apply the sum rule. Really.

    You can find that sin(a+b) =sin(a)cos(b)+cos(a)sin(b)

    The function n=sin((A+B)/2)/sin(A/2) simplifies to

    [itex]n=\frac{\sin(A/2) \cos(B/2) + \cos(A/2) \sin(B/2)}{\sin(A/2)}=\cos(B/2)+\sin(B/2) \cot(A/2)[/itex].

    When squaring, you can remove the half-angles by using the identities

    cos2(A/2)=0.5(1+cos(A)), which you have tried, but did something wrong. cos^2((A+D)/2)+sin^2((A+D)/2)=1 instead that complicated something you wrote.

    By the way, you replaced cos(A/2)/sin(A/2) with arctan(A/2), which is wrong. It is cotangent, cot(A/2).

  7. Sep 18, 2011 #6

    OK, so from my original formula:


    [tex]\implies n=\cos(\frac{D}{2})+\cot(\frac{A}{2})\sin(\frac{D}{2})[/tex]

    [tex]\implies\frac{\partial n}{\partial A}=[-\frac{1}{\sin^{2}(\frac{A}{2})}](\frac{1}{2})\sin(\frac{D}{2})=-\frac{1}{2}\frac{\sin(\frac{D}{2})}{\sin^{2}(\frac{A}{2})}[/tex]

    [tex]\implies|\frac{\partial n}{\partial A}|^{2}=\frac{1}{4}[\frac{\sin(\frac{D}{2})}{\sin^{2}(\frac{A}{2})}]^{2}[/tex]

    and for:

    [tex]\frac{\partial n}{\partial D}[/tex]

    [tex]\frac{\partial n}{\partial D}=-(\frac{1}{2})\sin(\frac{D}{2})+(\frac{1}{2})\cot(\frac{A}{2})\cos(\frac{D}{2})[/tex]


    [tex]\implies|\frac{\partial n}{\partial D}|^{2}=(\frac{1}{4})\sin^{2}(\frac{D}{2})-(\frac{1}{2})\frac{\sin(\frac{D}{2})\cos(\frac{D}{2})}{\tan(\frac{A}{2})}+(\frac{1}{4})\frac{\cos^{2}(\frac{D}{2})}{\tan^{2}(\frac{A}{2})}[/tex]



    Can you please confirm that this is correct? (although I still used the chain rule... I don't think that sum rule would have made life much easier, but stand to be corrected there...)

    please note, this formula for n is for finding refractive index of material using a prism where A and D are two angles.... the reason I want square of partial derivatives is so that I can then use them in the error propogation formula along with my error in A and D in order to find error in n. So, I suppose it is not essential that I reduce the trig. to the most simple form possible so long as the final formula is in some way tractable - that said, dealing with trig. has been a weakness of mine for a very long time and so I am eager to improve in this area.
  8. Sep 18, 2011 #7


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    The derivative with respect to D is simpler in the original form,
    [tex]\frac{\partial n}{\partial D}=\frac{1}{2}\frac{\cos(\frac{A+D}{2})}{sin(\frac{A}{2})}[/tex]. The derivative with respect to A is
    [tex]\frac{\partial n}{\partial A}=-\frac{1}{2}\frac{\sin(\frac{D}{2})}{\sin^{2}(\frac {A}{2})}[/tex]
    You need not do the squares. When evaluating the squares of the derivatives, the simplest way is to evaluate the derivatives and then square them (the numbers, not the formulae).

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