⁵Since $c=m-a$, maximizing over $c$ is equivalent to maximizing over $a$. Differentiating $u(m-a)+v\succ (a)$ with respect to $a$ gives $-u^{\partial }(c)+v{\succ }^{\partial }(a)=0$, which rearranges to $u^{\partial }(c)=v{\succ }^{\partial }(a)$.