Could someone explains what's gone wrong here?

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Firepanda
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I just read this on another forum

'We know that the derivative of x^2 with respect to x is 2x. However, what if we rewrite x^2 as the sum of x x's, and then take the derivative:

d/dx[ x^2 ] = d/dx[ x + x + x + ... (x times) ]
= d/dx[x] + d/dx[x] + d/dx[x] ... (x times)
= 1 + 1 + 1 + ... (x times)
= x

This argument shows that the derivative of x^2 with respect to x is actually x. So what's going on here?'

Wheres the mistake? :P
 
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What happens if x is not a positive integer? What if x is negative or a fraction like 1/2? How would it evaluate?

However, that's not all that's wrong with this argument. Are you familiar with the definition of the derivative? If you are, then take a closer look at the definition to see if this method is still valid under the original fundamental definition.
 
The function
f(x)= 5x= (5+ 5\c+ 5 \cdot\cdot\cdot+ 5){x times} <br /> is a function of x: your method would say <br /> f'(x)= (0+ 0+ 0\cdot\cdot\cdot+ 0)= 0 <br /> which is, of course, not true. <br /> f(x)= x^2= (x+ x+ \cdot\cdot\cdot+ x) {x times} <br /> <br /> is a function of x in <b>two</b> ways. You have not taken into account the "x times" part of the function.
 
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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