# 2nd derivative change of variables

#### BrokenPhysics

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?

#### Samy_A

Homework Helper
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?

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:
$\displaystyle \frac{d^2f}{dy^2}=\frac{d}{dx}(\frac{df}{dy})\frac{dx}{dy}=\frac{d}{dx}(\frac1{\sqrt a}\frac{df}{dx})\frac{dx}{dy}=\frac1{\sqrt a}\frac{d²f}{dx²}\frac1{\sqrt a}=\frac1{ a}\frac{d²f}{dx²}$

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#### BrokenPhysics

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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