# Is this differentiation correct?

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: 172

HallsofIvy
Homework Helper
No, it isn't.

SteamKing
Staff Emeritus
Homework Helper
What happens to the exponent when we differentiate?

No, it isn't.

Are you sure?

What happens to the exponent when we differentiate?

dI/dS = (- N U/ D2 ) / ( S U/ D + 1)2

SteamKing
Staff Emeritus
Homework Helper
That looks better.

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.

Mark44
Mentor

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.

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?

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! ^_^

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
101 KB · Views: 150
Mark44
Mentor
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:
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
14.7 KB · Views: 155
Last edited:
Mark44
Mentor
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

Got it. Thank you Mark44. With that correction, it is clear to me now.
:thumbs: