# Simple Differentiation - Is this legal / correct method?

1. Oct 1, 2015

### Sirsh

I have the equation:
x = 2*L*sin(θ/2) and in my lecture notes they have converted it to: ϑx = L*cos(θ/2)*ϑθ

Is it correct to do the following to get this answer?

x = 2*L*sin(θ/2)
x = 2*L*sin(θ/2)*(ϑ(θ/2)/ϑx)
x*ϑx = 2*L*sin(θ/2)*ϑ(θ/2)
1*ϑx = (1/2)*2*L*cos(θ/2)*ϑθ
ϑx = L*cos(θ/2)*ϑθ

My problem is I don't see how you can keep ϑx and ϑθ after the differentiation operation has been done, and if it is correct to be able to separate ϑ(θ/2) into (1/2)*ϑθ ?

Any help would be appreciated, Thanks.

2. Oct 1, 2015

### SteamKing

Staff Emeritus
It's called implicit differentiation:

http://www.sosmath.com/calculus/diff/der05/der05.html

This approach is a little more long-winded, but it's good you get the same result.
You are treating dx and dθ as differential quantities here. Doing d(θ/2) = (1/2)dθ is perfectly acceptable.

Haven't you studied integration by parts or the use of u-substitution to solve integrals yet?

3. Oct 2, 2015

### Staff: Mentor

I'm not sure what character you used, but the above should be dx = L * cos(θ/2) dθ
No. Differentiate both sides with respect to x.
From this you get 1 = 2 * L cos(θ/2) * d/dx(θ/2), or
1 = 2 * L cos(θ/2) * 1/2 * dθ/dx

You can then solve algebraically for dθ/dx.
You can work with differentials: d(θ/2) = (1/2) dθ
The chain rule is in play here. $d(θ/2) = \frac{d(θ/2)}{dθ} \cdot dθ = (1/2) dθ$