שינויים
/* שאלה 6 */
<math>(arctan(\frac{lnx}{1+x^2}))'=\frac{(\frac{lnx}{1+x^2})'}{1+(\frac{lnx}{1+x^2})^2}=\frac{\frac{\frac{1+x^2}{x}-2xlnx}{(1+x^2)^2}}{1+(\frac{lnx}{1+x^2})^2}=\frac{\frac{1+x^2-4x^2lnx}{x(1+x^2)^2}}{\frac{(x^2+1)^2+(lnx)^2}{(1+x^2)^2}}=\frac{1+x^2-4x^2lnx}{x(x^2+1)^2+x(lnx)^2}</math>
'''ג.'''
<math>(x^{-n}+n^{-x})'=-nx^{-n-1}-lnn\cdot n^{-x}</math>
