examday

/TMUA

Consider the false claim $\dfrac{\mathrm{d}(x^2)}{\mathrm{d}x}=x$.

Below is an attempt at the proof.

• 1. We can write $x^2$ as $x+x+x+\ldots$ ($x$ times) for any $x$
• 2. Given (1), we have $\dfrac{\mathrm{d}(x^2)}{\mathrm{d}x}=\dfrac{\mathrm{d}}{\mathrm{d}x}(x+x+x+x+\ldots)$ ($x$ times)
• 3. Further, $\dfrac{\mathrm{d}}{\mathrm{d}x}(x+x+x+\ldots)$ ($x$ times) is $\dfrac{\mathrm{d}(x)}{\mathrm{d}x}+\dfrac{\mathrm{d}(x)}{\mathrm{d}x}+\dfrac{\mathrm{d}(x)}{\mathrm{d}x}\ldots$ ($x$ times)
• 4. And $\dfrac{\mathrm{d}(x)}{\mathrm{d}x}+\dfrac{\mathrm{d}(x)}{\mathrm{d}x}+\dfrac{\mathrm{d}(x)}{\mathrm{d}x}\ldots$ ($x$ times) simplifies to $1+1+1+1\ldots$ ($x$ times)
• 5. $\ldots$ which is equal to $x$

Which of the above steps is/are wrong?