Prove me wrong !!

[sakibtanvir]

We know that,\[\frac{\partial }{\partial x}(x^2)=2x
\]We also know that,\[x^2=(x+x+x+x+x....+x)
\]($x$ times).Now,\[\frac{\partial }{\partial x}(x^2)
\]
\[=\frac{\partial }{\partial x}(x+x+x+x...x)
\]
\[=(1+1+1+....+1)
\]\[=x
\]






Where is the crime

[Phlembac Adib Hasan]

Look, you are finding derivative for real $x$s.But you can write $x^2=x+x+x...+x\; \; $ ($x$ times) only for integer $x$s.If x is not integer, you have to write $x^2=x+x+...+x\; \; (\left \lfloor x \right \rfloor-1) \; \; times\; \; +${$x$}$x$.So you can not say that.This is the mistake,I think.

[sm.joty]

Look, you are finding derivative for real $x$s.But you can write $x^2=x+x+x...+x\; \; $ ($x$ times) only for integer $x$s.If x is not integer, you have to write $x^2=x+x+...+x\; \; (\left \lfloor x \right \rfloor-1) \; \; times\; \; +${$x$}$x$.So you can not say that.This is the mistake,I think.

No the actual fact is not that. Note that you here don't consider the fact of constant and variable. Shakib here treated the variable as a constant. But note that $x^2=(x+x+x+x+x....+x)$ is a function of $x$ thus you can't take it as a constant.

For details see this link.

http://www.gonitpathshala.org/%E0%A6%95 ... %E0%A6%BE/