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2nd derivative change of variables

  1. Feb 18, 2016 #1
    Let's say ##f(x)=ax^2##. Then ##d^2f/dx^2=2a##.

    Now we can make the change of variables ##y\equiv\sqrt ax## to give ##f(y)=y^2##. Then ##d^2f/dy^2=2##.

    It follows that
    ##\frac{d^2f}{dx^2}=a\frac{d^2f}{dy^2},##
    but I can't replicate this with the chain rule.

    I would put
    ##\frac{df}{dy}=\frac{df}{dx}\frac{dx}{dy}=\frac1{\sqrt a}\frac{df}{dx}##
    ##\frac{d^2f}{dy^2}=\frac{d^2f}{dx^2}\frac{dx}{dy}+\frac{df}{dx}\underbrace{\frac{d^2x}{dy^2}}_0=\frac1{\sqrt a}\frac{d^2f}{dx^2},##

    which is a factor of ##1/\sqrt a## different from what we know the answer should be. So what am I doing wrong?

    Thanks in advance!
     
  2. jcsd
  3. Feb 18, 2016 #2

    Samy_A

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    This step is wrong:
    ##\frac{d^2f}{dy^2}=\frac{d^2f}{dx^2}\frac{dx}{dy}+\frac{df}{dx}\frac{d^2x}{dy^2}##
    By the chain rule, it should be:
    ##\displaystyle \frac{d^2f}{dy^2}=\frac{d}{dx}(\frac{df}{dy})\frac{dx}{dy}=\frac{d}{dx}(\frac1{\sqrt a}\frac{df}{dx})\frac{dx}{dy}=\frac1{\sqrt a}\frac{d²f}{dx²}\frac1{\sqrt a}=\frac1{ a}\frac{d²f}{dx²}##
     
    Last edited: Feb 18, 2016
  4. Feb 18, 2016 #3
    That makes sense. Thanks a lot!
     
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