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I'm learning derivative by my own, can someone help please?

The question is: Using the definition of the derivative, across the boundary, find the derivative of the function $f(x) = \sin (x)$.

thanks

M47145
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Bunny
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  • Assuming you are starting from scratch, the hard work is in establishing some inequalities on $\sin h,\cos h$ for small $h$, which you can do by elementary geometry. Then you just use the formula $\sin(x+h)=\sin x\cos h+\cos x\sin h$. – almagest Apr 18 '16 at 20:54
  • Have you set anything up? If I recall, you need a standard trig identity then there's one key limit in the way, after that. – pjs36 Apr 18 '16 at 20:55
  • It also depends on how you have defined $\sin(x)$ in the first place. In some contexts, it is preferred to define $\sin(x)$ as its taylor expansion or to define $\sin(x)$ using complex exponentials. $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. In either of these alternate definitions, the derivative follows from rules of deriving polynomials or deriving exponentials and the corresponding related definition of $\cos(x)$. – JMoravitz Apr 18 '16 at 21:13

2 Answers2

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You will need two things. First that $\sin(x+h)=\sin x\cos h + \sin h\cos x$, and secondly that $$\lim_{\theta \to 0}\frac{\sin\theta}{\theta}=1$$ (beautiful proof here). In fact you'll need to use this last identity to show that $$\lim_{\theta \to 0}\frac{\cos \theta-1}{\theta}=0$$ So three things. It might not be the absolute shortest proof but I'm fond of it.

Community
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K.Power
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    +1 Yes, I agree that is the approach with the least number of pre-requisites. – almagest Apr 18 '16 at 21:17
  • I'm glad you agree thanks for the upvote. No matter the proof you're going to need the first limit which is the only part that requires any real ingenuity. I'm actually going to share a link to that proof for the OP. – K.Power Apr 18 '16 at 21:21
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We have $$\lim_{h\rightarrow 0}\frac{\sin(x+h)-\sin x}{h}=\lim_{h\rightarrow 0}\frac{2\cos(x+\frac h2) \sin\frac h2}{h}=\\\lim_{h\rightarrow 0}\frac{\sin\frac h2}{\frac h2}\lim_{h\rightarrow 0}\cos (x+\frac h2)=1*\cos x=\cos x $$ as desired.

You just have to remember a few trigonometric properties and work your way with the algebra involved.

MathematicianByMistake
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