pochoda
: 19 sty 2010, 20:32
\(10arctg \frac{x}{2}-x\)
\((10arctg \frac{x}{2}-x)' = (10arctg \frac{x}{2})' - x' = 10(arctg \frac{x}{2})' -1 = 10(\frac {1}{1+(\frac{x}{2})^2})(\frac{x}{2})' -1=10(\frac {1}{1+(\frac{x}{2})^2})*\frac{1}{2}-1=\)
\(=5\(\frac {1}{1+\(\frac{x}{2}\)^2}\)-1=\(\frac {5}{1+(\frac{x}{2})^2}\)-1\)
chyba tak to leci...