derivative of $\frac 2x \sin(x^3)$ by definition The function: $$\frac 2x \sin(x^3)$$ when $x\neq0$, and $0$ while $x=0$.
I need to find if the function is derivation at $x=1$.
First step was to check if the function is continuous. there is just 1 side of limit to check (since its the same function around $x=1$, so I compared the limit of the function with $f(1)$:
$$\lim_{x\to 1}\space \frac{2}{x}\sin(x^3) = 2\sin(1) = f(1)$$
first question: was this step necessary?
next:
$$f'(1) =\lim_{h\to 0}\space \frac{ \frac2{h+1}\sin((h+1)^3)-2\sin(1)}h$$
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3)}{h+1}-2\sin(1)}h$$
and I know about $\lim_{x\to 0}\frac {\sin(x)}x = 1$, so:
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3) \cdot (h+1)^2}{(h+1)^3}-2\sin(1)}h$$
and then:
$$=\lim_{h\to 0}\space \frac{ 2(h+1)^2-2\sin(1)}h$$
and now I don't know how to get rid of the $h$ denominator

For the first question, no, you don't have to show its continuity at that point. Just show that the limit $$ f'(1) = 2\lim_{h\to 0} \frac{\frac{\sin (h + 1)^3}{h + 1} - \sin 1}{h}$$
exists. To compute the limit, you can use
\begin{align*}
f'(1) &= 2\lim_{h\to 0} \frac{\sin (h + 1)^3 - \sin 1 - h\sin 1}{h(h + 1)}\\
&= 2 \left(\lim_{h\to 0} \frac{\sin (h + 1)^3 - \sin 1}{h(h + 1)} - \sin 1\right)
\end{align*}
you can ignore the $h + 1$ in the denomininator since it approaches $1$. Hence $$ f'(1) = 2\left(\lim_{h\to 0}\frac{\sin(h + 1)^3 - \sin 1}{h}\right) - 2\sin 1$$
the first term can be compute using $\sin a - \sin b$ formula, in fact, it is the derivarative of $\sin x^3$ at $x = 1$.
