Which Equation Has the Greatest Rate of Change?

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Homework Statement



I believe it is no?
I'm asked which equation has the greatest rate of change, first derivative, going back to the deffinition of derivative I would say yes...

Well there are two equations I calculated the first derivative in one to be 2 and the other one to be -2... so if the derivative is a vector which my gut tells me it is then 2 would be the correct answer becasue it's greater in value than -2 but if the derivative is a scalar and only has direction the both answers have the greatest...

I also have a hunch that technically speaking right the derivative has no width right and so therefore is neither a scalar or a vector so...

please help which one has the greatest rate of change the one with the first derivative of 2 or -2... I don't know if it's a vector or not...

THANK YOU!

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The Attempt at a Solution

 
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Given an equation dependent on only one variable (say x), the derivative evaluated at a point for that equation is a scalar value, not a vector.

The rate of change they are asking for seems to be "direction" independent and so in your case the two functions have equal rates of change.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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