Math
Name:
Anonymous
2008-05-29 16:45
Whats y´ of y=x^2/x+1
Name:
Anonymous
2008-06-20 21:37
\TeX
Name:
Anonymous
2008-06-20 21:38
\frac{5}{\frac{5}{5}}
Name:
Anonymous
2008-06-20 21:39
\frac{5}{\frac{5}{\frac{5}{\frac{5}{\frac{5}{\frac{5}{\frac{5}{\frac{5}{}}}}}}}}
Name:
RedCream
2008-06-21 0:25
Sadly, this inclusion of TEX in /sci/ only makes up for 1% of the damage that mootle did when he completely fagged up /n/.
Name:
Sweden
2008-06-21 4:33
f(x) = \frac{x^2}{x} + 1
Easy approach:
f(x) = x + 1
Which obviously leads to
f'(x) = x^0 + 1 = 2
Different approach:
f(x) = x^2 \cdot x^{-1}
f'(x) = 2x \cdot x^{-1} + x^2 \cdot (-x^{-2}) = \frac{2x}{x} - \frac{x^2}{x^2} = x
Name:
Sweden
2008-06-21 4:34
Seems I forgot something above:
-------------------
f(x) = \frac{x^2}{x} + 1
Easy approach:
f(x) = x + 1
Which obviously leads to
f'(x) = x^0 + 1 = 2
Different approach:
f(x) = x^2 \cdot x^{-1}
f'(x) = 2x \cdot x^{-1} + x^2 \cdot (-x^{-2}) = \frac{2x}{x} - \frac{x^2}{x^2} = x
Name:
Anonymous
2008-06-21 4:37
f(x) = \frac{x^2}{x} + 1
Easy approach:
f(x) = x + 1
Which obviously leads to
f'(x) = x^0 + 1 = 2
Different approach:
f(x) = x^2 \cdot x^{-1}
f'(x) = 2x \cdot x^{-1} + x^2 \cdot (-x^{-2}) = \frac{2x}{x} - \frac{x^2}{x^2} = x
Name:
Anonymous
2008-06-21 23:21
>>84
MrVacBob did the patch, not moot
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