2nd derivative change of variables

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!
 

Samy_A

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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!
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:
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}##
That makes sense. Thanks a lot!
 

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