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Explanation for this derivative

  1. Oct 20, 2012 #1
    The derivative of sin(ax²) is 2axcos(ax²). Why is that? I understand the cos(ax²) part of the derivative but I don't understand where the 2ax is coming from.
     
  2. jcsd
  3. Oct 20, 2012 #2

    MarneMath

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    Do you know the chain rule?
     
  4. Oct 20, 2012 #3

    You can google "Chain rule". It says that

    $$\left(f(g(x)\right)'=f'(g(x))\cdot g'(x)$$

    In the present case we have [itex]\,f(x)=\sin x\,\,\,,\,\,\,g(x)=ax^2\,[/itex]

    DonAntonio
     
  5. Oct 20, 2012 #4
    that makes a lot more sense. thanks
     
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