Finding Derivative of sinx: What do I do Next?

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Discussion Overview

The discussion revolves around finding the derivative of the sine function, specifically addressing the steps to take after deriving an expression involving sine and hyperbolic functions. The scope includes mathematical reasoning and technical explanation related to calculus.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant states that the derivative of sin(x) is cos(x) but expresses uncertainty about the next steps in the calculation.
  • Another participant emphasizes the importance of limits in the calculation and provides two specific limits that need to be evaluated.
  • A different participant suggests simplifying the expression sin(x)cosh + cosh(sin(x)) into a single trigonometric function before proceeding with further calculations.
  • One participant reiterates the suggestion to simplify the expression and acknowledges the help received from another participant, while also mentioning a lack of familiarity with LaTeX notation.

Areas of Agreement / Disagreement

Participants express varying approaches to the problem, with no consensus on a single method to proceed after obtaining the derivative. Multiple competing views remain regarding the simplification and evaluation of limits.

Contextual Notes

Some participants reference specific limits and simplifications without fully resolving the mathematical steps involved. There is an assumption that certain limits are known, which may not be universally understood.

iRaid
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I know the derivative is cosx but I don't know what to do.
So I get the derivative of sinx down to:
[tex]\frac{sinxcosh}{h} - \frac{sinx}{h} + \frac{cosxsinh}{h}[/tex]

What do I do after that?
(not a homework problem)
 
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Hi iRaid! :smile:

First about notation: don't forget the limit!

I see that you need to calculate two limits now:

[tex]\lim_{h\rightarrow 0}{\frac{\cos(h)-1}{h}}=0~~\text{and}~~\lim_{h\rightarrow 0}{\frac{\sin(h)}{h}}=1[/tex]

The second limit should be known to you (if not, see http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus for a nice derivation).
The first limit follows from the second by multiplying numerator and denominator by 1+cos(x).
 
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You should simplify sinxcosh+coshsinx into one simple trig function and then use the http://www.sosmath.com/trig/prodform/prodform.html" .
 
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rock.freak667 said:
You should simplify sinxcosh+coshsinx into one simple trig function and then use the http://www.sosmath.com/trig/prodform/prodform.html" .

Well that makes it much easier also @micromass, thanks too :) and I didn't forget the limit, just I didn't know how to do it in latex :p
 
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