Derivative of $\sqrt[5]{\sin(x)}$ from first principles. I'm a newbie to derivatives using first principle. I've just learnt how to differentiate basic functions using first principles.
My problem is that, how can we differentiate $\sqrt[4]{\sin x}$ or $\sqrt[5]{\sin(x)}$.
I'm able to find the derivative of $\sqrt{\sin x}, \sqrt[3]{\sin{x}}$ but I found the fourth root and 5th root somewhat difficult.
First of all, I thought that I could find derivative of $\sqrt[4]{\sin x}$ using the identity $(A- B)=\dfrac{A^4 - B^4}{(A+B)(A^2+B ^2)}$ considering $A = \sqrt[4]{\sin (x+h)}, B =\sqrt[4]{\sin (x)} $. 
So far so good.
But I think it's not a good way to solve it. Like if we have $\sqrt[9]{\sin x}$, we cannot make identities.
I'm wondering if there is more interesting way to evaluate $\lim\limits_{h\to0}\dfrac{\sqrt[n]{\sin (x+h)} - \sqrt[n]{\sin(x)}}{h}$.
 A: You can also use
$$\frac{(1+\alpha)^k-1}{\alpha}\to k\tag{1}\label{1}$$
when $\alpha\to 0$, in this way
\begin{eqnarray}
\mathcal L &=& \lim_{h \to 0} \frac{\sqrt[n]{\sin(x+h)} - \sqrt[n]{\sin x}}{h}=\\
&=&\lim_{h\to 0}\frac{\sqrt[n]{\sin x \cos h + \cos x \sin h} - \sqrt[n]{\sin x}}{h}=\\
&=&\sqrt[n]{\sin x}\lim_{h\to 0}\frac{\sqrt[n]{\cos h +\frac{\cos x \sin h}{\sin x}}-1}{h}=\\
&=&\sqrt[n]{\sin x} \lim_{h\to 0}\frac{\sqrt[n]{1+\left(\cos h-1 +\frac{\cos x \sin h}{\sin x}\right)}-1}{h}=\\
&\stackrel{\eqref{1}}{=}&\frac{\sqrt[n]{\sin x}}n\lim_{h\to 0}\left[\underbrace{\frac{\cos h - 1}{h}}_{\to 0}+\underbrace{\frac{\sin h}{h}}_{\to 1}\cdot \frac{\cos x}{\sin x}\right]=\\
&=& \frac1n \frac{\cos x}{\sqrt[n]{\sin^{n-1} x}}
\end{eqnarray}
A: Let's assume,
$$ y= \sqrt[5]{\sin(x)}$$
Using the definition of first principles,
$$\begin{aligned}y' &= \lim_{h\to0}\dfrac{ \sqrt[5]{\sin(x + h)} -  \sqrt[5]{\sin(x)} }{h}\\y' &= \lim_{h\to0}\dfrac{ \sqrt[5]{\sin(x + h)} -  \sqrt[5]{\sin(x)} }{h}\cdot \dfrac{\sin(x+h) - \sin(x)}{\sin(x+h) - \sin(x)}\\y' &= \lim_{h\to0}\dfrac{ \sqrt[5]{\sin(x + h)} -  \sqrt[5]{\sin(x)} }{\sin(x+h) - \sin(x)}\cdot \dfrac{\sin(x+h) - \sin(x)}{h}\\y' &= \lim_{h\to0}\dfrac{ \sqrt[5]{\sin(x + h)} -  \sqrt[5]{\sin(x)} }{\sin(x+h) - \sin(x)}\cdot \lim_{h\to0}\dfrac{\sin(x+h) - \sin(x)}{h}\end{aligned}$$
For the first limit, apply the following identity:
$$\boxed{\lim_{x\to 0}\dfrac{x^{n} - a^n}{x - a} = n \ x^{n-1}}$$
For the second limit, apply trig. identity.
$$\boxed{\sin(A) - \sin(B) = 2\cos\left(\frac{A+B}{2}\right)\cdot \sin\left(\frac{A-B}{2}\right)}$$
Using these,
$$\begin{aligned}&y' = \dfrac{1}{5}\sin^{1/5 -1}(x)\cdot \lim_{h\to0}\dfrac{2\cos(x + h/2)\cdot \sin(h/2)}{h}\\&y' = \dfrac{1}{5}\sin^{-4/5 }(x)\cdot\cos(x) \lim_{h\to0}\dfrac{\sin(h/2)}{h/2}\\& y' = \dfrac{1}{5}\sin^{-4/5 }(x)\cdot\cos(x) (1)\\& \underline{y' = \dfrac{1}{5}\sin^{-4/5 }(x)\cdot\cos(x)}  \end{aligned}$$

$n$th root of $\sin(x)$ can also be derived using same procedure.
