Understanding Derivatives: The Role of the Chain Rule Explained | Homework Help

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Homework Statement


I attached a problem from my homework.

My only issue is that I don't know why it is x arc sec and not 2x arc sec etc...shouldn't there be a 2x from the chain rule from the derivative inside the square root? I cannot understand for my life why there is an x and not a 2x.

I kept getting this wrong until I googled it and someone in yahoo answers gave a response without a 2, so i tried it and my online homework said it was right. But don't know why!


Homework Equations





The Attempt at a Solution

 

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Well, I didn't get the same result, maybe I have to simplify it, but the 2 factor goes away because the 1/2 exponent "comes down" when you derive sqrt (x^2 - 1). Is that what you are talking about? And why are you talking about arcsec?
 
Oh dear. That was a silly mistake! I definitely didn't bring down the 1/2 when I took the derivative. ::slaps forehead:: It's been a while since I took calc 1 so I am making a lot of really silly mistakes now in Calc 2!

I couldn't see for my life what I was missing!

Oh and I was saying arcsec because i thought sec^-1x was more confusing to read.

Thank you for your help!
 
1/sec = cos btw
 
Well

(\sec^{-1} x)' = \frac{1}{x\sqrt{x^2 -1}}

so that explains the first term. As for the second, there's no 2, because of the chain rule.
 
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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