How to compute derivative of $\sin(x^3)$ by definition? I am trying to proof that derivative of $\sin(x^3) = 3x^2\cos(x^3)$ by definition.
But I don't know an identity for $\sin(x^3)$ for getting $\cos(x^3)$.
Even I try to find a quantity similar to $\frac{\sin(x)}{x}$.

$$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0} \frac{\sin(x+h)^3-\sin x^3}{h}$$
$(x+h)^3=x^3(1+h/x)^3$ Let us use $(1+z)^k=1+kz+\frac{k(k-1)}{2} z^2+O(z^3)$ to get
$$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0}\frac{\sin[x^3(1+h/x)^3]-\sin x^3}{h}$$
$$=\lim_{h\rightarrow 0}\frac{\sin[x^3+3hx^2+O(h^2)]-\sin x^3}{h}$$
$$=\lim_{h\rightarrow 0} \frac{\sin x^3\cos(3hx^2)+\cos x^3 \sin(3hx^2)-\sin x^3 }{h}$$
$$=\lim_{h\to 0} \frac{\sin x^3[(\cos 3hx^2)-1]}{h}+\cos x^3 \lim_{h\to 0}\frac {\sin 3hx^2}{h}$$
Use $\cos z=1-z^2/2+O(z^4), \sin z=z+O(z^3)$, then we get
$$\frac{d\sin x^3}{dx}=\sin x^3\lim_{h\to 0} \frac{9h^2x^4}{h}+\cos x^3 \lim_{h\to 0}\frac{3hx^2}{h}=0+3x^2 \cos x^3$$
