Why Does the Chain Rule Apply Differently to the Derivative of (x^2+x)^(1/2)?

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D= (x^2+x)^(1/2)

Why does the chain rule in this case produce D' = (1/2)(x^2+x)^(-1/2) (2x+1)(x')

and not D' = (1/2)(x^2+x)^(-1/2) (2x+1)(2x')
 
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\frac{dD}{dt}=\frac{dx}{dt} \times \frac{dD}{dx}

So like in the other thread, can you understand why now?
 
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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