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How is this differentiated

  1. Feb 22, 2016 #1
    image.jpg
    I cannot understand the first part (second line) where it says dw/dk
    Can someone explain in a more step by step method
     
  2. jcsd
  3. Feb 22, 2016 #2

    fresh_42

    Staff: Mentor

    If ##f(x) = \sqrt {ax^2+b}## then ##\frac{df}{dx} = \frac{d}{dx} (ax^2+b)^{\frac{1}{2}}## and by chain rule $$ \frac{df}{dx} = \frac{1}{2} (ax^2+b)^{- \frac{1}{2}} \cdot \frac{d}{dx} (ax^2+b) = \frac{1}{2} \cdot \frac{1}{(ax^2+b)^{\frac{1}{2}}} \cdot (2ax) = \frac{1}{2} \cdot \frac{2ax}{\sqrt{ax^2+b}}.$$ Now substitute ##x=k##, ##b=m^2c^4## and ##a=ħ^2c^2##.
     
  4. Feb 23, 2016 #3

    Mark44

    Staff: Mentor

    @lioric, this is pretty straightforward stuff,usually taught in the first semester/quarter of calculus. You need to go back and review differentiation topics, especially the chain rule, which can be used to calculate the derivative shown here.
     
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