How will we solve this differentiation? I am not not understanding from where to start. $\frac{Root 3}{4}*a^2 *2 + 3a$ $\frac{4}{root 3* a^2}$
We have solve dA/da [That whole equation is equal to A]
What I am not getting is that from which function to start. Like there is already root present and then there is a^2.
What I did
I solved the first part using product rule. u=root 3*4 and v = 4.
If we simplify $A = \frac {\sqrt 3}{4}(a^2)(2) + (3a)\frac 4{(\sqrt 3)(a^2)},$ we get $A = \frac {\sqrt 3}{2}a^2 + \frac {4\sqrt 3}{a}.$
Now differentiate. Each term is pretty simple, just apply the power rule.
$\frac {dA}{da} = \sqrt 3a - \frac {4\sqrt 3}{a^2}$
The $\sqrt 3$ is just a constant, and should be treated like any other constant. It just rides along.