Answer
$m^{\prime}(t)=-6(5t^{4}-1)^{3}(85t^{4}-1)$
Work Step by Step
Use the product rule
$\color{blue}{[u(t)\cdot v(t)]^{\prime}=u(t)\cdot v^{\prime}(t)+v(t)\cdot u^{\prime}(t)}$
For $v^{\prime}(t)$, we need the chain rule.
$\left[\begin{array}{llll}
u(t)=-6t & , & v(t)=(5t^{4}-1)^{4} & \\
u^{\prime}(t)=-6 & & v^{\prime}(t)=[w(t)]^{4} & (\text{chain rule})\\
& & v^{\prime}(t)=4\cdot[w(t)]^{4}\cdot w^{\prime}(t) & \\
& & v^{\prime}(t)=4(5t^{4}-1)^{3}\cdot(20t^{3}) &
\end{array}\right]$
$m^{\prime}(t)=(-6t)[(5t^{4}-1)^{4}]^{\prime}+(5t^{4}-1)^{4}(-6t)^{\prime}$
$=-6t[4(5t^{4}-1)^{3}\cdot 20t^{3}]+(5t^{4}-1)^{4}(-6)$
$=-480t^{4}(5t^{4}-1)^{3}-6(5t^{4}-1)^{4}$
$=-6(5t^{4}-1)^{3}[80t^{4}+(5t^{4}-1)]$
$=-6(5t^{4}-1)^{3}(85t^{4}-1)$
