A Differentiation Dilemma
Posted: Thu Jan 17, 2013 8:59 pm
Let, $f(x)=x^2$.
So, $\displaystyle f'(x)=\frac{\mathrm {d}}{\mathrm {d}x} \big( f(x) \big)=\frac{\mathrm d}{\mathrm dx}(x^2)=2x$.
Again,
\[\displaystyle f(x)=x^2=x\times x=\underbrace{x+x+x+\ldots+x}_{x\,\text{times}}\]
Therefore,
\[\displaystyle f'(x)=\frac{\mathrm {d}}{\mathrm{d}x}\big(f(x)\big)\]
\[=\frac{\mathrm{d}}{\mathrm{d}x}\big(\underbrace{x+x+x+\ldots+x}_{x\,\text{times}}\big)\]
\[=\underbrace{\frac{\mathrm{d}}{\mathrm{d}x}(x)+\frac{\mathrm{d}}{\mathrm{d}x}(x)+\frac{\mathrm{d}}{\mathrm{d}x}(x)+\ldots+\frac{\mathrm{d}}{\mathrm{d}x}(x)}_{x\,\text{times}}\]
\[=\underbrace{1+1+1+\ldots+1}_{x\,\text{times}}=x\,.....\text{!!!} \]
What is wrong in this solution......???
Source:"অলিম্পিয়াড সমগ্র".
So, $\displaystyle f'(x)=\frac{\mathrm {d}}{\mathrm {d}x} \big( f(x) \big)=\frac{\mathrm d}{\mathrm dx}(x^2)=2x$.
Again,
\[\displaystyle f(x)=x^2=x\times x=\underbrace{x+x+x+\ldots+x}_{x\,\text{times}}\]
Therefore,
\[\displaystyle f'(x)=\frac{\mathrm {d}}{\mathrm{d}x}\big(f(x)\big)\]
\[=\frac{\mathrm{d}}{\mathrm{d}x}\big(\underbrace{x+x+x+\ldots+x}_{x\,\text{times}}\big)\]
\[=\underbrace{\frac{\mathrm{d}}{\mathrm{d}x}(x)+\frac{\mathrm{d}}{\mathrm{d}x}(x)+\frac{\mathrm{d}}{\mathrm{d}x}(x)+\ldots+\frac{\mathrm{d}}{\mathrm{d}x}(x)}_{x\,\text{times}}\]
\[=\underbrace{1+1+1+\ldots+1}_{x\,\text{times}}=x\,.....\text{!!!} \]
What is wrong in this solution......???
Source:"অলিম্পিয়াড সমগ্র".