Problem in understanding chain rule: second derivative

Question

I have the next functions

\begin{align} &z(u,v)\\ &u=xy\\ &v=x²+y²\end{align}

and I want to get the second derivative with respect to $x$.

I calculate the first derivative:

$$z_{x} = z_{u}\cdot u_{x}+z_{v}\cdot v_{x} = z_{u} \cdot y +z_v\cdot 2x$$

I calculate the second derivative:

\begin{align} (z_{x})_{x} &= (z_{x})_{u}\cdot u_{x}+(z_{x})_{v}\cdot v_{x}\\ &=(z_{u}\cdot y+z_{v}\cdot 2x)_{u}\cdot u_{x}+(z_{u}\cdot y+z_{v}\cdot 2x)_{v}\cdot v_{x} \\ &=(z_{uu}\cdot y+z_{vu}\cdot 2x)\cdot y+(z_{uv}\cdot y+z_{vv}\cdot 2x)\cdot 2x\\ &=z_{uu}\cdot y^2+z_{u}\cdot 2xy+z_{v}\cdot 2xy+z_{vv}\cdot 4x^2 \end{align}

Where is the error? The teacher put $2z_{v}$ in $(z_{x})_{x} = (z_{x})_{u}\cdot u_{x}+(z_{x})_{v}\cdot v_{x} + 2z_{v}$

Why?

Answer 1

In the last step of your calculation, you seem to have simplified $z_{uv}$ and $z_{vu}$ into $z_v$ and $z_u$. This is incorrect.

If $z$ is sufficiently smooth (it suffices for its second partial derivatives to be continuous), then $z_{uv}=z_{vu}$ and the final result should be

$$z_{uu}\cdot y^2+z_{uv}\cdot 4xy+z_{vv}\cdot 4x^2$$

I'm not quite sure what your teacher did.

Answer 2

Your work: $$\begin{align} (z_{x})_{x} &= (z_{x})_{u}\cdot u_{x}+(z_{x})_{v}\cdot v_{x}&&\checkmark\text{chain rule}\\ &=(z_{u}\cdot y+z_{v}\cdot 2x)_{u}\cdot u_{x}+(z_{u}\cdot y+z_{v}\cdot 2x)_{v}\cdot v_{x}&&\checkmark\text{chain rule} \\ &=(z_{uu}\cdot y+z_{vu}\cdot 2x)\cdot y+(z_{uv}\cdot y+z_{vv}\cdot 2x)\cdot 2x&&{\times\text{did not apply the product rule}\\(z_v\cdot 2x)_u=z_{vu}\cdot 2x+z_v\cdot(2x)_u}\end{align}$$

Your teacher's work:

$$\begin{align}(z_x)_x &= (z_u\cdot y+z_v\cdot 2x)_x&&\text{chain rule}\\&=(z_u)_x\cdot y+(z_v)_x\cdot 2x+z_v\cdot 2 &&\text{product rule}\\&= (z_{uu}\cdot y+z_{uv}\cdot 2x)\cdot y+(z_{vu}\cdot y+z_{vv}\cdot 2x)\cdot 2x+z_v\cdot 2&&\text{chain rule}\\&=z_{uu}\cdot y^2+z_{uv}\cdot 4xy+z_{vv}\cdot 4x^2+z_v\cdot 2 \end{align} $$

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Note: order of application is important too.