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
 A: Note that this step:
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3) * (h+1)^2}{(h+1)^3}-2\sin(1)}h$$
and then:
$$=\lim_{h\to 0}\space \frac{ 2(h+1)^2-2\sin(1)}h$$
is not correct since you are solving the limit in two parts and this not allowed.
Moreover:
$$\frac{\sin(h+1)^3 }{(h+1)^3}\to \sin1$$
A: I did not answer your exact question, I looked at the case of $x=0$ rather than $x=1$, but the argument is analogous. 
Evaluate the derivative of $f : \mathbb{R} \to \mathbb{R}$, where $$f(x) : = \begin{cases}
\frac{2}{x} \sin(x^3), & x \in \mathbb{R} \backslash \{ 0 \}, \\
0, & \text{otherwise}.
\end{cases}$$ 
Let us first verify that $f$ is at least continuous at $x=0$, let alone differentiable. To this end, write $\sin(x)$ as its Taylor expansion $$\sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1}.$$ Then $$\frac{2}{x} \sin(x^3) = 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{3(2k+1) - 1} = 2\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{6k+2}. $$ We therefore see that $$\lim_{x \to 0} \frac{2}{x} \sin(x^3) =0,$$ verifying that $f$ is continuous at $0$. Now let us check whether $f$ is differentiable at $x=0$. Indeed, we observe that \begin{eqnarray*}
\lim_{\Delta x \to 0} \frac{f(0+\Delta x) - f(0)}{\Delta x} &=& \lim_{\Delta x \to 0} \frac{f(\Delta x) - f(0)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{2}{\Delta x \cdot \Delta x} \sin ( (\Delta x)^3 ). 
\end{eqnarray*} For simplicity, let $z = \Delta x$ and note that we just need to determine whether the limit $$\lim_{z \to 0} \frac{2 \sin(z^3)}{z^2}$$ exists. Analogously to the above continuity argument, write $\sin(z^3)$ as its Taylor expansion, $$\sin(z^3) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{3(2k+1)}.$$ It then follows that \begin{eqnarray*}
\lim_{z \to 0} \frac{2}{z^2} \sin(z^3) &=& \lim_{z \to 0} 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{6k+2} \\
&=& 0,
\end{eqnarray*} so $f$ is differentiable with $f'(0)=0$. 
A: The function is obviously continuous at $1$, because it is a composition of continuous functions.
Your computation is flawed: it is true that $\lim_{x\to0}\frac{\sin x}{x}=1$, but you have $\lim_{h\to0}\frac{\sin((1+h)^3)}{(1+h)^3}$ and the preceding limit doesn't apply.
For the derivative at $1$:
$$
f'(1)=\lim_{h\to0}\frac{\dfrac{2\sin((1+h)^3)}{1+h}-2\sin1}{h}
$$
Now write $(1+h)^3=1+hg(h)$, where $g(h)=3+3h+h^2$, so you have to compute, disregarding the factor $2$,
$$
\frac{\sin(1+hg(h))-(1+h)\sin 1}{h(1+h)}=
\frac{\sin1\cos(hg(h))+\cos1\sin(hg(h))-\sin 1-h\sin 1}{h(1+h)}
$$
The right hand side can be rewritten as
$$
\frac{\cos(hg(h))-1}{hg(h)}\frac{g(h)\sin 1}{1+h}
+\frac{g(h)\cos 1}{1+h}\frac{\sin(hg(h))}{hg(h)}
-\frac{\sin 1}{1+h}
$$
Now the limit is basically done; the first fraction has limit $0$, the second fraction has limit $3\sin 1$; the third fraction has limit $3\cos 1$, the fourth fraction has limit $1$; the fifth fraction has limit $\sin 1$. Thus, reinserting the factor $2$:
$$
f'(1)=2(3\cos1-\sin1)
$$
Just to check with the standard procedure: for $x\ne0$,
$$
f'(x)=2\frac{3x^3\cos(x^3)-\sin(x^3)}{x^2}
$$
and $f'(1)=2(3\cos 1-\sin 1)$.
Are you sure you don't want to compute the derivative at $0$? In order to do it, you need to first check the function is continous at $0$:
$$
\lim_{x\to0}\frac{2\sin(x^3)}{x}=
\lim_{x\to0}2x^2\frac{\sin(x^3)}{x^3}=0
$$
Then
$$
f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}
=\lim_{h\to0}\frac{2\sin(h^3)}{h^2}
=\lim_{h\to0}2h\frac{\sin(h^3)}{h^3}
$$
A: 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 $x = 1$.
A: 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.
