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

When you have
$$\frac{\frac{2 \sin \left((h+1)^3\right)}{h+1}-2 \sin (1)}{h}$$
you must rewrite for all $h\neq0$ to
$$\frac{\sin \left(\frac{1}{2} \left((h+1)^3-1\right)\right)}{\frac{1}{2} \left((h+1)^3-1\right)}\frac{2 \left(h^2+3 h+3\right) \cos \left(\frac{1}{2} \left((h+1)^3-1\right)+1\right)}{h+1}-\frac{2 \sin (1)}{h+1}$$
and then take the limit by substituting $h =0$ except for the first fraction which has known limit equal to $1$.
When rewriting you simply has to follow the proof for the derivative of a product, a composition, and sine.
