Is this differentiation correct?

In summary, Mark44 states that 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}##
  • #1
musicgold
304
19
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.
 

Attachments

  • diff question Sept 2013.pdf
    21.9 KB · Views: 238
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  • #2
No, it isn't.
 
  • #3
What happens to the exponent when we differentiate?
 
  • #4
HallsofIvy said:
No, it isn't.

Are you sure?
 
  • #5
SteamKing said:
What happens to the exponent when we differentiate?

Oh! How about this?

dI/dS = (- N U/ D2 ) / ( S U/ D + 1)2
 
  • #6
That looks better.
 
  • #7
SteamKing said:
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
musicgold said:
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
Mark44 said:
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
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
ModestyKing said:
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?
 

Attachments

  • dI dS chart.pdf
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  • #12
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
Mark44 said:
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.
 

Attachments

  • DI DS calc1.xlsx
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Last edited:
  • #14
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
Got it. Thank you Mark44. With that correction, it is clear to me now.
:thumbs:
 

FAQ: Is this differentiation correct?

What is differentiation?

Differentiation is a mathematical process used to find the rate at which one variable changes with respect to another variable. It is a fundamental concept in calculus and is used to solve problems involving rates of change, optimization, and related rates.

How is differentiation done?

Differentiation is done by taking the derivative of a function. This involves finding the slope of the function at a specific point by calculating the limit of the function as the change in the independent variable approaches zero. The resulting derivative represents the instantaneous rate of change of the function at that point.

What is the purpose of differentiation?

The purpose of differentiation is to analyze and understand the behavior of functions. It allows us to find the maximum and minimum values of a function, determine the direction in which the function is increasing or decreasing, and solve optimization problems. Differentiation is also used in physics, engineering, and other fields to model and solve real-world problems.

Is differentiation always correct?

In theory, differentiation is always correct as it follows a set of rules and principles. However, it is possible to make mistakes when applying these rules, resulting in an incorrect derivative. It is important to double-check the work and use different methods, such as graphing, to confirm the result.

Can differentiation be used for any function?

Differentiation can be used for most functions, including polynomial, exponential, logarithmic, and trigonometric functions. However, there are some functions that are not differentiable, such as discontinuous or non-continuous functions. In these cases, other methods, such as numerical approximation, may be used.

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