Answer Happy

\begin{tabular}{|l|l|l} f′(x)=x3−6x2−15x+20 thru (0,−2) & f′(x)=x​x2+x+1​ thru (0,−3) \\ \hlinef′(x)=4+3x4​ thru (0,2) & \\ & f′(x)=2cos(2x) thru (42π​,1) \\ \hline \end{tabular}