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Why derivative of sin x is cos x, only if x is in radians?

  1. Apr 14, 2013 #1
    What differente does it make? As far as I can see it, the limit definition of a derivative shouldn't be affected by the fact that x is expressed in radians or degrees...
     
  2. jcsd
  3. Apr 15, 2013 #2
    The limit definition of the derivative relies heavily on approximations to length. The radian has 2 meanings, that of an angle and that of a length so calculus works very nicely when angles are represented in radians.

    :smile:
     
  4. Apr 15, 2013 #3
    You can think of it mechanically in terms of the chain rule. If the nice derivative rule holds for radians, it can't hold for an angle measure that differs from radians by a factor.
    If f(y)=siny, where y is in degrees, then f(y)=sin(yπ/180), where y is in degrees but the argument of the sine is now in radians so we know how to take its derivative. This has a derivative of π/180*cos(πy/180). The logic here is the same as why, for example [itex] \frac {d}{dx} e^x = e^x [/itex] but [itex] \frac {d}{dx} e^{2x} \neq e^{2x} [/itex]
     
    Last edited: Apr 15, 2013
  5. Apr 15, 2013 #4
    The limit [itex] \lim_{x→0}\frac{sin x}{x} = 1 [/itex] holds only when x is in radians. This limit is essential to differentiating sin(x) from first principles.

    BiP
     
  6. Apr 15, 2013 #5
    Thank you so much for your answers, now I see it clearly :)
     
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