Partial derivative of a square root

  • #1
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Hi,

I'm using partial derivatives to calculate propagation of error. However, a bit rusty on my calculus.

I'm trying to figure out the partial derivative with respect to L of the equation:

2pi*sqrt(L/g)

(Yep, period of a pendulum). "g" is assumed to have no error. I know I can use the chain rule...

so, 2pi*(L/g)^(1/2) --> 2pi*1/2*(L/g)^(-1/2) , or pi*(L/g)^(-1/2).

I am doing this correctly? Or did I just take the derivative (and not the partial derivative)?

Thanks!
 

Answers and Replies

  • #2
SteamKing
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Hi,

I'm using partial derivatives to calculate propagation of error. However, a bit rusty on my calculus.

I'm trying to figure out the partial derivative with respect to L of the equation:

2pi*sqrt(L/g)

(Yep, period of a pendulum). "g" is assumed to have no error. I know I can use the chain rule...

so, 2pi*(L/g)^(1/2) --> 2pi*1/2*(L/g)^(-1/2) , or pi*(L/g)^(-1/2).

I am doing this correctly? Or did I just take the derivative (and not the partial derivative)?

Thanks!
Partial derivative of what w.r.t. L? You haven't provided an equation, just an expression.
 
  • #3
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2pi*sqrt(L/g) = T, which is a function of L.
 
  • #4
SteamKing
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Rewrite as T = 2π*(L/g)1/2. Does that give you any ideas? Remember, g is a constant.
 
  • #5
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Using the chain rule, I can bring down the 1/2 and subtract 1 from the exponent, so

dL/dT = 1/2*2π*(L/g)-1/2 or dL/dT = π*(L/g)-1/2

Though, now it seems that I'm not treating "g" as a constant.
 
  • #6
SteamKing
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Using the chain rule, I can bring down the 1/2 and subtract 1 from the exponent, so

dL/dT = 1/2*2π*(L/g)-1/2 or dL/dT = π*(L/g)-1/2

Though, now it seems that I'm not treating "g" as a constant.
Well, factor g out of the square root before taking the derivative.

Technically, you are not using the chain rule. You are using the power rule.
 
  • #7
Fredrik
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Using the chain rule, I can bring down the 1/2 and subtract 1 from the exponent, so

dL/dT = 1/2*2π*(L/g)-1/2 or dL/dT = π*(L/g)-1/2

Though, now it seems that I'm not treating "g" as a constant.
Don't you mean dT/dL? As SteamKing said, you can rewrite the original expression with g outside the square root before you differentiate it. If you don't, you will have to use the chain rule. It tells us that there's an extra factor that you didn't include in the quote above.
 

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