12.19 Given the function $y=x^{4}-6 x^{2}+1$. Find the maximum and minimum values of its derivative on the interval $[-1,3]$.


12.19 We have $y^{\prime}=4 x^{3}-12 x=f(x)$. We find the values of the function $f(x)$ at the ends of the interval $[-1,3]: f(-1)=8, f(3)=72$. Now let's find the critical points of the function $f(x)$ that belong to this interval: $f^{\prime}(x)=y^{\prime \prime}=$ $=12 x^{2}-12=12\left(x^{2}-1\right) ; f^{\prime}(x)=0$, if $x= \pm 1$. Then we get $f(-1)=8$ (this value was found earlier), $f(1)=-8$. Thus, $y_{\text {min }}^{\prime}=-8, y_{\text {max }}^{\prime}=72$.

Answer: $y_{\text {min }}^{\prime}=-8, y_{\text {max }}^{\prime}=72$.