Simple Differentiation - Is this legal / correct method?

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The discussion centers on the application of implicit differentiation to the equation x = 2*L*sin(θ/2) and its conversion to ϑx = L*cos(θ/2)*ϑθ. Participants confirm that differentiating both sides with respect to x is valid, and that the separation of ϑ(θ/2) into (1/2)*ϑθ is acceptable due to the chain rule. The method discussed is a legitimate approach to implicit differentiation, affirming that ϑx and ϑθ can be retained after differentiation.

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Sirsh
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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.
 
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Sirsh said:
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
It's called implicit differentiation:

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

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
This approach is a little more long-winded, but it's good you get the same result.
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.
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?
 
Sirsh said:
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
I'm not sure what character you used, but the above should be dx = L * cos(θ/2) dθ
Sirsh said:
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)
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.
Sirsh said:
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 ?
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##
 

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