Can you explain in detail please how to find the derivative of this function? $$\sin^3(312x^2)$$
The derivative of $f(g(x))$ is (if it exists) $$f'(g(x))g'(x)\tag1$$ (application of chain rule).
Here $g(x)=312x^2$ (can you find $g'(x)$?).
Note that we can write $f(x)=h(k(x))$ where $h(x)=x^3$ and $k(x)=\sin x$ so that according to $(1)$: $$f'(x)=h'(k(x))k'(x)$$
This must be enough.
$$y = \sin^3(312x^2)$$
This is a composite function. Whenever you have a composite function, use the Chain Rule.
$$(f\circ g)’(x) = f’(g(x))\cdot g’x$$
Let $f(x) = x^3$ and $g(x) = \sin(312x^2)$.
$$\implies y’ = 3\sin^2(312x^2)\cdot [\sin (312x^2)]’$$
Here, you have to apply the same technique again to simplify the second part.
$$[\sin(312x^2)]’ = \sin’(312x^2)\cdot (312x^2)’ = \cos(312x^2)\cdot 64x = 624x\cos(312x^2)$$
So, the expression becomes
$$y’ = 3\sin^2(312x^2)\cdot 624x\cos(312x^2) = 1872x\sin^2(312x^2)\cos(312x^2)$$