0
$\begingroup$

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 .

  • 0
    What's first principle of derivative? – 2017-01-20
  • 0
    @MyGlasses Finding the derivative by first principles – 2017-01-20
  • 0
    Please search in google . I can't write the formula here for technical problem . Sorry . – 2017-01-20
  • 0
    What do you mean by too long? Do you have the feeling that your solution may be wrong? If that is the case you should probably post your result and we can help. – 2017-01-20
  • 0
    No .Actually I am searching a rather shorter method . But that too without using chain rule . – 2017-01-20
  • 0
    $$f'(x) =\lim_{h\to0} \frac{f(x+h) - f (x)}{h}$$ – 2017-01-20

3 Answers 3

2

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

This is $0$ for $k

So if we collect all summands, we get $$\frac d {dx} \sin^n(x) = n \sin(x)^{n-1}\cos(x).$$

  • 0
    I know you are interested in a short solution, but this is probably the shortest one if you don't want to use the chain or product rule. – 2017-01-20
  • 0
    Ok...good...thank you . – 2017-01-20
  • 0
    did you expect something different? – 2017-01-20
0

Differentiate it as a composite function: $$f'(x)=n(\sin x)^{n-1}\cos x.$$

  • 1
    Can we have a solution without chain rule ? – 2017-01-20
  • 0
    If you want: $$(f_1f_2\dots f_n)'=f_1'f_2\dots f_n+f_1f_2'\dots f_n+\dots+f_1f_2\dots f_n'.$$ Then write $\sin^n x$ as a product of $n$ factors. – 2017-01-20
  • 0
    Ok..good idea . I did not think about the generalisation of uv rule. – 2017-01-20
  • 1
    The OP is looking for a method of differentiation via first principles. – 2017-01-20
  • 1
    What do you mean by O P ? – 2017-01-20
  • 1
    @GopeshPatgiri OP is short for Original Poster. I.e. the OP is you. – 2017-01-20
  • 0
    @Bacon, OP wrote my idea was good. I don't understand your comment. – 2017-01-20
  • 0
    @szw1710 The question states to use first principle of differentiation. You have not. You have used the chain rule. – 2017-01-20
  • 0
    Ok...thanks...though I was looking for that method yet I found the above method helpful . Anyway any more idea is welcome . – 2017-01-20
0

Why not replicate the proof of chain rule. We need to use two standard limits: $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1},\,\lim_{x\to 0}\frac{\sin x} {x} =1\tag{1}$$ If $f(x) = \sin^{n} x$ then \begin{align} f'(x) &=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\notag\\ &=\lim_{h\to 0}\frac{\sin^{n}(x+h)-\sin^{n}x}{h}\notag\\ &=\lim_{h\to 0}\frac{\sin^{n}(x+h)-\sin^{n}x}{\sin(x+h)-\sin x} \cdot\frac{\sin (x+h) - \sin x} {h} \notag\\ &=\lim_{t\to u} \frac{t^{n} - u^{n}} {t-u} \cdot\lim_{h\to 0}\frac{\sin(x+h)-\sin x} {h} \notag\\ &=nu^{n-1}\lim_{v\to 0}\frac{\sin(x+2v)-\sin x} {2v}\notag\\ &=nu^{n-1}\lim_{v\to 0}\frac{2\cos(x+v)\sin v} {2v}\notag\\ &=nu^{n-1}\lim_{v\to 0}\cos(x+v)\cdot\frac{\sin v} {v} \notag\\ &=nu^{n-1}\cos x\notag\\ &=n\sin^{n-1}x\cos x\notag \end{align} In the above derivation we have used the substitutions $$t=\sin(x+h), u= \sin x, h=2v$$ and used the continuity of $\sin x, \cos x$.