How can we find the derivative of the function $f$ defined by
$$f(x)=\sin^n(x)$$
by first principle of derivative, i.e. by calculating the limit
$$f'(x)= \lim_{h\to 0} \frac {f(x+h)-f(x)} h.$$
I used binomial theorem but it is too long .
[Math] Derivative of $\sin^n(x)$ by first principle
calculus
Best Answer
Assuming you mean "directly from the definition" by "from first principles", I can provide an answer for $n\in \mathbb{N}$.
We need two intermediate results which I will not prove: $$\lim_{h\to 0} \sin(h)/h =1, \qquad \lim_{h\to 0} \sin^k(h)/h =0, ~k>1,$$ $$\lim_{h\to 0}\frac{1-\cos(h)^n}h=0, \qquad n>0,$$ $$\sin(x+h)=\sin(x) \cos(h)+\cos(x)\sin(h).$$
The derivaive of $\sin^n$ at $x$ is defined as $$\frac d {dx} \sin^n(x)=\lim_{h\to 0} \frac 1 h \left(\sin^n(x+h)-\sin^n(x)\right).$$ Using the binomial theorem, we are interested in the limit of terms of the form $$\lim_{h\to 0}\frac{\left(\sum_{k=0}^n {n \choose k }\sin(x)^k\cos(x)^{n-k} \cos(h)^k \sin(h)^{n-k}\right) -\sin^n(x)}h,\qquad k=0,\dots n-1.$$
Due to the linearity of the limit we can investigate each power of $\sin(h)$ separately.
If $k=n$, then $$\lim_{h\to 0}\frac{\sin(x)^{n} \cos(h)^n -\sin^n(x)}h=\sin^n(x) \lim_{h\to 0}\frac{1-\cos(h)^n}h=0.$$ For $k<n$ we have $${n \choose k}\lim_{h\to 0} \frac{\sin(x)^k\cos(x)^{n-k} \cos(h)^k \sin(h)^{n-k}}h={n \choose k} \lim_{h\to 0} \sin(x)^k\cos(x)^{n-k} \frac{ \sin(h)^{n-k}}h.$$
This is $0$ for $k<n-1$ and for $k=n-1$: $$\dots={n \choose n-1} \sin(x)^{n-1}\cos(x) \lim_{h\to 0} \frac{ \sin(h)}h=n \sin(x)^{n-1}\cos(x).$$
So if we collect all summands, we get $$\frac d {dx} \sin^n(x) = n \sin(x)^{n-1}\cos(x).$$