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 $\frac{\sin(x)}{x}$.
 A: $$\begin{align}
\frac{d}{dx}\sin(x^3)&=\lim_{h\to0}\frac{\sin{((x+h)^3)-\sin (x^3)}}{h}\\
\\
&=\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)\cos \left(\frac{(x+h)^3+x^3}{2}\right)}}{h}\\
\\
&=\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\cdot\lim_{h\to0}\cos \left(\frac{(x+h)^3+x^3}{2}\right)\\
\\
&=\left(\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\right)\cdot\cos \left(\lim_{h\to0}\frac{(x+h)^3+x^3}{2}\right)\\
\\
&=\left(\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\right)\cdot\cos \left(x^3\right)\\
\\
&=\cos \left(x^3\right)\cdot\lim_{h\to0}\left(\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\cdot\frac{\frac{(x+h)^3-x^3}{2}}{\frac{(x+h)^3-x^3}{2}}\right)\\
\\
&=\cos \left(x^3\right)\cdot\lim_{h\to0}\frac{\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{\frac{(x+h)^3-x^3}{2}}\cdot\lim_{h\to0}\frac{(x+h)^3-x^3}{h}\\
\\
&=\cos \left(x^3\right)\cdot 1\cdot\lim_{h\to0}\frac{(x+h)^3-x^3}{h}\\
\\
&=\cos \left(x^3\right)\cdot 3x^2\\
\end{align}$$
A: 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 $f'(a)=3a^2\cos(a^3)$.
A: $$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0} \frac{\sin(x+h)^3-\sin x^3}{h}$$
$(x+h)^3=x^3(1+h/x)^3$ 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
$$\frac{d\sin x^3}{dx}=\sin x^3\lim_{h\to 0} \frac{9h^2x^4}{h}+\cos x^3 \lim_{h\to 0}\frac{3hx^2}{h}=0+3x^2 \cos x^3$$
