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 $\lim\limits_{h\to0}\dfrac{\sqrt[n]{\sin (x+h)} – \sqrt[n]{\sin(x)}}{h}$.
Best Answer
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}