Review in Variable Differentiation

jgreen520
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Please see attached.

I was looking for an explanation of the answer I have attached. Its been a little while and was just looking for the logic behind the differentiation shown for this problem. Its basically an optimization problem where I am looking for the minimum angle (theta) for the least amount of force for T_ab.

Thanks
 

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It uses the standard rule for differentiating a quotient: d(u/v) = (vdu - udv)/v2
 
You can also do the differentiation of a quotient as the differentiation of a product using the chain rule: u/v= uv^{-1} so that
(u/v)'= (uv^{-1})'= u'v^{-1}+ u(v^{-1})'= u'v^{-1}+ u(-v^{-2}v')
= \frac{u'}{v}+ \frac{uv'}{v^2}= \frac{u'v}{v^2}+ \frac{uv'}{v^2}= \frac{u'v+ uv'}{v^2}
 
HallsofIvy said:
= \frac{u'}{v}+ \frac{uv'}{v^2}= \frac{u'v}{v^2}+ \frac{uv'}{v^2}= \frac{u'v+ uv'}{v^2}
= \frac{u'}{v}- \frac{uv'}{v^2} etc.
 
Thanks for the response.

So differentiating D-LCos(theta) the constant D goes to 0 and the -LCos(theta) goes to +Lsin(theta) ...then multiply that by sin(theta) in the denominator to get Lsin^2(theta)...

However I was curious why nothing happens to the L...in front of the Cos.

Thanks
 
"L" is a constant. What do you think should happen to it?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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