Answer
False
Work Step by Step
By Definition 9.7.3, the sixth-degree Taylor polynomial about $x = {x_0}$ for $f$ is
${p_6}\left( x \right) = f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + \dfrac{{f{\rm{''}}\left( {{x_0}} \right)}}{{2!}}{\left( {x - {x_0}} \right)^2} + \dfrac{{f{\rm{'''}}\left( {{x_0}} \right)}}{{3!}}{\left( {x - {x_0}} \right)^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( 6 \right)}}\left( {{x_0}} \right)}}{{6!}}{\left( {x - {x_0}} \right)^6}$
Taking the derivative gives
$p_6^{\left( 1 \right)}\left( x \right) = f'\left( {{x_0}} \right) + f{\rm{''}}\left( {{x_0}} \right)\left( {x - {x_0}} \right) + 3\dfrac{{f{\rm{'''}}\left( {{x_0}} \right)}}{{3!}}{\left( {x - {x_0}} \right)^2} + \cdot\cdot\cdot + 6\dfrac{{{f^{\left( 6 \right)}}\left( {{x_0}} \right)}}{{6!}}{\left( {x - {x_0}} \right)^5}$
$p_6^{\left( 1 \right)}\left( {{x_0}} \right) = f'\left( {{x_0}} \right)$
Continuing with taking the derivatives and so on, we obtain $p_6^{\left( 4 \right)}\left( {{x_0}} \right) = {f^{\left( 4 \right)}}\left( {{x_0}} \right)$.
Therefore, the statement is false.