<s>[INST] 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 $\frac{\sin(x)}{x}$.
 [/INST] Computation using the differentiability of $\sin$ on $\mathbb R$
Let $f(x)=\sin(x^3)$. Then, for every $a\in\mathbb R$ and $x\neq a$, we have
$$
\frac{f(x)-f(a)}{x-a} 
= \frac{\sin(x^3)-\sin(a^3)}{x-a} 
= \frac{\sin(x^3)-\sin(a^3)}{x^3-a^3} \frac{x^3-a^3}{x-a} 
$$
Let $X=x^3$ and $A=a^3$. Then $\frac{\sin(x^3)-\sin(a^3)}{x^3-a^3}=\frac{\sin(X)-\sin(A)}{X-A}$ and since $\sin$ is differentiable at $A$ and $X\to A$ when $x\to a$, we get
$$\lim_{x\to a}\frac{\sin(x^3)-\sin(a^3)}{x^3-a^3} = \lim_{X\to A}\frac{\sin(X)-\sin(A)}{X-A} = \cos(A) = \cos(a^3)$$
Moreover, $\frac{x^3-a^3}{x-a}=x^2+ax+a^2 \to 3a^2$ as $x\to a$. By Limit Laws, we have
$$
\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=3a^2\cos(a^3) 
$$
This means that $f$ is differentiable at $a$ and $f'(a)=3a^2\cos(a^3)$.
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