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How to calculate Derivative of sin sq. root x by definition?

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate derivative of (sin sq. root x) w.r.t x?

    2. Relevant equations

    Limit Δx--> 0 (sin√(x+Δx) - sin(√x)) / Δx

    3. The attempt at a solution

    i couldn't operate it from here.... Δy = (2cos((√x+Δx) + (√x)) . sin((√x+Δx) - (√x)) / Δx...???
     
  2. jcsd
  3. Sep 19, 2013 #2

    arildno

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    Rewrite:
    [tex]\sin(\sqrt{x+Dx})=\sin(\frac{\sqrt{x}}{\sqrt{1+\frac{Dx}{\sqrt{x}}}}
    +\frac{\frac{Dx}{\sqrt{x}}}{\sqrt{1+\frac{Dx}{\sqrt{x}}}})[/tex]
     
    Last edited: Sep 19, 2013
  4. Sep 19, 2013 #3
    where did that sq. root (1+ Dx/sq.root x) come from?
     
  5. Sep 19, 2013 #4

    arildno

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    Multiply the argument within the sine as follows:
    [tex]\sqrt{x+Dx}=\sqrt{x+Dx}*1=(\sqrt{x+Dx})*\frac{\sqrt{x+Dx}}{\sqrt{x+Dx}}=\frac{x+Dx}{\sqrt{x+Dx}}[/tex]
    Now, extraxt "x" from the square root in the denominator and simplify.
     
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