- #1
BrokenPhysics
- 4
- 0
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!
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!