Is this differentiation correct?

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  • #1
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Hi,

Would you be able to tell me if my differentiation in the attached file correct?
Note that N, U and D are constants.

I am trying to understand how I changes with a change in S.

Thanks.
 

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  • #2
HallsofIvy
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No, it isn't.
 
  • #3
SteamKing
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What happens to the exponent when we differentiate?
 
  • #4
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  • #5
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What happens to the exponent when we differentiate?
Oh! How about this?

dI/dS = (- N U/ D2 ) / ( S U/ D + 1)2
 
  • #6
SteamKing
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That looks better.
 
  • #7
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That looks better.
Thanks SteamKing. I am still struggling with one question:
The reason behind this calculation was to understand the sensitivity of I to S. I wanted to know how much I changes for a 1% change in in S. It seems the differentiation doesn't answer that.

What should I do to understand the sensitivity?

Thanks.
 
  • #8
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Oh! How about this?

dI/dS = (- N U/ D2 ) / ( S U/ D + 1)2
It's much simpler to write the equation in this form:
I = N(US + D)-1

Differentiating with respect to S yields
dI/dS = -N * (US + D)-2 * U = ##\frac{-NU}{(US + D)^2}##

Also, to find the sensitivity of I to small changes in S, write the equation above using differentials.


dI = ##\frac{-NU}{(US + D)^2} dS ##

For small changes in S, you can approximate dS by ΔS and dI by ΔI.
 
  • #9
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Also, to find the sensitivity of I to small changes in S, write the equation above using differentials.


dI = ##\frac{-NU}{(US + D)^2} dS ##

For small changes in S, you can approximate dS by ΔS and dI by ΔI.
How do I use this equation to find the change in I for a 1% change in S?
 
  • #10
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What Mark is saying (I believe) is that if you replace dI with change in I (the answer you want to get) and dS with change in S (which you have; 1% = 0.01), then you're set. The concept is this: dI/dS is the change of dI over dS; the change of output compared to input. So, in dI/dS, set dS to what it is; the change in input, or 0.01. Now multiply both sides by 0.01 to get dI = (rest of equation). There, you've got a ratio! ^_^
 
  • #11
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What Mark is saying (I believe) is that if you replace dI with change in I (the answer you want to get) and dS with change in S (which you have; 1% = 0.01), then you're set. The concept is this: dI/dS is the change of dI over dS; the change of output compared to input. So, in dI/dS, set dS to what it is; the change in input, or 0.01. Now multiply both sides by 0.01 to get dI = (rest of equation). There, you've got a ratio! ^_^
I am still a bit confused. Please see the attached file.
N=100, U= 110, S = 9 and D=120

With these values I = 9.01%
Also these values give me dI/dS = -0.009

Now, for dS = 1%, I get dI= -0.01%. However, this result doesn't make any sense.
For example, if I actually increase S by 1% from 9.00 to 9.09, I changes from 9.01% to 8.93%, a decline of -0.08%, much higher than the 0.01% indicated above.

What am I missing?
 

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  • #12
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You didn't show what you did to get your value of dI, so it's hard to say what you're missing. Using the formula I wrote in post #8, I get dI ≈ -8.93 X 10-5.

I used the values you show in the PDF:
N = 100
U =110
S = 9
D = 120
ΔS = .01
So dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .01##

Edit: Corrections to the above
When S increases by 1%, that means that ΔS = .09, not .01.
This results in a value for dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .09 ≈ -8.035 X 10^{-4}##
 
Last edited:
  • #13
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You didn't show what you did to get your value of dI, so it's hard to say what you're missing. Using the formula I wrote in post #8, I get dI ≈ -8.93 X 10-5.
Sorry about that. You are right, that is how I also calculated it dI= -0.0000893
See the attached Excel file.
 

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  • #14
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Why are you converting to percentages? I think they are throwing you off.

If S increases by 1%, ΔS = .09, so the corrected value for ΔI is about -.0008035.

Calculating I directly with S = 9 results in .09009 (approx.)
Calculating I directly with S = 9.09 results in .08929

So I has decreased (making ΔI negative), and by direct calculation we see that ΔI = -.00080. This agrees reasonably well with the value obtained using differentials.

Using the values I showed in post 12, I get dI ≈ -0.0000893

Calculating I directly with S = 9, I get I ≈ .09009
Increasing S by 1% (to 9.09), I get I ≈ .089293
 
  • #15
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Got it. Thank you Mark44. With that correction, it is clear to me now.
:thumbs:
 

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