== שאלה 6 ==
גזירות:'''א.''' <math>(x^{cos(e^{x^2})})'=(e^{cos(e^{x^2})lnx})'=e^{cos(e^{x^2})lnx}\cdot (cos(e^{x^2})lnx)'=x^{cos(e^{x^2})}(-2xln(x)e^{x^2}sin(e^{x^2})+\frac{cos(e^{x^2})}{x})</math> '''ב.''' <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>
