Rate of change of a cubes area in respect to the space diagonal The space diagonal of a cube shrinks with $0.02\rm m/s$. How fast is the area shrinking when the space diagonal is $0.8\rm m$ long?

I try:
Space Diagonal = $s_d = \sqrt{a^2+b^2+c^2}=\sqrt{3a^2}$ Where $a$ is the length of one side.
  Area = $a^2$
Rate of change for $s_d$ with respect to $a$
$${\mathrm d\over \mathrm da}s_d={\mathrm d\over \mathrm da}\sqrt{3a^2}={\sqrt{3}a \over \sqrt{a^2}}$$
Rate of change for $\rm area$ with respect to $a$
$${\mathrm d\over \mathrm da}\mathrm{area}={\mathrm d\over \mathrm da}a^2={2a}$$
Im stuck when it comes to calculating one thing from another thing! However I have no problem when it comes to position, velocity and acceleration! Can anybody solve this? 

The simplest way is to express the area $a$ of the cube as a function of the length $d$ of the space diagonal. Given $d$ the side length $s$ of the cube is
$$s={1\over\sqrt{3}} \ d\ ,$$
and the total surface area $a$ then becomes
$$a=6s^2=2 d^2\ .$$
Now all quantities appearing here are in fact functions of $t$; therefore at all times $t$ we have
$$a(t)=2d^2(t)\ .$$
It follows that
$$a'(t)=4d (t) \ d'(t)\ ,$$
and the data $d(t_0)=0.8$ m, $d'(t_0)=-0.02$ m/sec imply that $a'(t_0)=-0.064$ m$^2$/sec.
