# Partial derivative of a square root

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!

SteamKing
Staff Emeritus
Homework Helper
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.

2pi*sqrt(L/g) = T, which is a function of L.

SteamKing
Staff Emeritus
Homework Helper
Rewrite as T = 2π*(L/g)1/2. Does that give you any ideas? Remember, g is a constant.

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.

SteamKing
Staff Emeritus
Homework Helper
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.

Fredrik
Staff Emeritus