Partial derivatives of a two-variable function with chain rule

calculusderivativesmultivariable-calculuspartial derivative

I have just started learning about partial derivatives and was wondering if I could get some confirmation as to whether my understanding is correct.

Here is the problem statement:

Suppose that $f(u, v)$ is a differentiable, two-variable function; its first-order partial derivatives are also differentiable. Let $z = f(u, v)$, $u = xy^2$, and $v = x^2$. Compute $\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial x}\right)$ using the chain rule.

Here is my thought process:

First we try to compute $\dfrac{\partial z}{\partial x}$. Applying the chain rule:

$$
\begin{align*}
\dfrac{\partial z}{\partial x} &= \dfrac{\partial}{\partial x} f(u, v) \\
&= \dfrac{\partial}{\partial u} f(u, v) \dfrac{\partial u}{\partial x} + \dfrac{\partial}{\partial v} f(u, v) \dfrac{\partial v}{\partial x} \\
&= f_u y^2 + f_v 2x
\end{align*}
$$

Next, partially differentiate the above with respect to $y$ to compute $\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial x}\right)$:

$$
\begin{align*}
\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial x}\right) &= \dfrac{\partial}{\partial y}\left(f_u y^2 + f_v 2x\right) \\
&= \dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial u} y^2\right) + \dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial v} 2x\right) \\
&= \dfrac{\partial}{\partial y}\dfrac{\partial z}{\partial u} y^2 + \dfrac{\partial z}{\partial u} 2y + \dfrac{\partial}{\partial y}\dfrac{\partial z}{\partial v} 2x + 0 \\
&= \dfrac{\partial}{\partial u}\dfrac{\partial z}{\partial y} y^2 + \dfrac{\partial z}{\partial u} 2y + \dfrac{\partial}{\partial v}\dfrac{\partial z}{\partial y} 2x
\end{align*}
$$

Computing $\dfrac{\partial z}{\partial y}$:

$$
\begin{align*}
\dfrac{\partial z}{\partial y} &= \dfrac{\partial z}{\partial u} \dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v} \dfrac{\partial v}{\partial y} \\
&= \dfrac{\partial z}{\partial u} 2xy + 0 \\
&= \dfrac{\partial z}{\partial u} 2xy
\end{align*}
$$

Plugging $\dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u} 2xy$ into the equation for $\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial x}\right)$:

$$
\begin{align*}
\dfrac{\partial}{\partial u}\dfrac{\partial z}{\partial y} y^2 + \dfrac{\partial z}{\partial u} 2y + \dfrac{\partial}{\partial v}\dfrac{\partial z}{\partial y} 2x &= \dfrac{\partial}{\partial u} \left(\dfrac{\partial z}{\partial u} 2xy\right) y^2 + \dfrac{\partial z}{\partial u} 2y + \dfrac{\partial}{\partial v}\left(\dfrac{\partial z}{\partial u} 2xy\right) 2x \\
&= f_{uu} 2xy^3 + f_u 2y + f_{uv} 4x^2y
\end{align*}
$$

Therefore, $\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial x}\right) = f_{uu} 2xy^3 + f_u 2y + f_{uv} 4x^2y$.

I have two big questions about this problem:

  1. Is my thought process correct? I feel like I may have misunderstood how the chain rule works with partial derivatives and would appreciate a sanity check for this.
  2. How do I compute the partial derivatives of $f$ with respect to $u$ and $v$? I have not been given any further information; can I assume that the first-order and second-order partial derivatives are in their simplest form?

Thank you.

Best Answer

Is my thought process correct? I feel like I may have misunderstood how the chain rule works with partial derivatives and would appreciate a sanity check for this.

The standard approach is chain rule, product rule, chain rule$\times 2$. Putting it all together you should obtain:

$$\small\begin{align} \dfrac{\partial^2 z}{\partial x~\partial y}&=\dfrac{\partial~~}{\partial y}\left[\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}\right] \\&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 u}{\partial x\,\partial y}+\dfrac{\partial~~}{\partial y}\left[\dfrac{\partial z}{\partial u}\right]\cdot\dfrac{\partial u}{\partial x}+\dfrac{\partial ~~}{\partial y}\left[\dfrac{\partial z}{\partial v}\right]\cdot\dfrac{\partial v}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 v }{\partial x\,\partial y} \\[1ex]&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 u}{\partial x\,\partial y}+\left[\dfrac{\partial^2 z }{\partial u~^2}\dfrac{\partial u}{\partial y}+\dfrac{\partial^2 z}{\partial u\,\partial v}\dfrac{\partial v}{\partial y}\right]\dfrac{\partial u}{\partial x}+\left[\dfrac{\partial^2 z}{\partial u\,\partial v }\dfrac{\partial u}{\partial y}+\dfrac{\partial^2 z}{\partial v~^2}\dfrac{\partial v}{\partial y}\right]\dfrac{\partial v}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 v }{\partial x\,\partial y} \\[1ex]&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 u}{\partial x\,\partial y}+\dfrac{\partial^2 z }{\partial u~^2}\dfrac{\partial u}{\partial y}\dfrac{\partial u}{\partial x}+\dfrac{\partial^2 z}{\partial u\,\partial v}\left[\dfrac{\partial v}{\partial y}\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial y}\right]+\dfrac{\partial^2 z}{\partial v~^2}\dfrac{\partial v}{\partial y}\dfrac{\partial v}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 v }{\partial x\,\partial y} \\[1ex]&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 xy^2}{\partial x\,\partial y}+\dfrac{\partial^2 z}{\partial u~^2}\dfrac{\partial xy^2}{\partial y}\dfrac{\partial xy^2}{\partial x}+\dfrac{\partial^2 z}{\partial u\,\partial v}\left[\dfrac{\partial x^2}{\partial y}\dfrac{\partial xy^2}{\partial x}+\dfrac{\partial x^2}{\partial x}\dfrac{\partial xy^2}{\partial y}\right]+\dfrac{\partial^2 z}{\partial v~^2}\dfrac{\partial x^2}{\partial y}\dfrac{\partial x^2}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 x^2 }{\partial x\,\partial y} \\[1ex]&=\dfrac{\partial z}{\partial u}2y+\dfrac{\partial^2 z }{\partial u~^2 }4xy^3+\dfrac{\partial^2 z}{\partial u\,\partial v}\left[0+4x^2y\right]+0+0 \\[1ex]&=2y f_u(xy^2,x^2)+2xy^3f_{uu}(xy^2,x^2)+4x^2yf_{uv}(xy^2,x^2)\end{align}$$

So...$\color{green}\checkmark$

Or more quickly: $$\begin{align}u&=xy^2\\ v&=x^2\\\dfrac{\partial^2 f(u,v)}{\partial x~\partial y} ~&=~\dfrac{\partial ~~}{\partial x}(\dfrac{\partial u}{\partial y}f_u(u,v)+\dfrac{\partial v}{\partial y} f_v(u,v)\\&=\dfrac{\partial~~}{\partial x}(2xyf_u(u,v)+0)\\&=\dfrac{\partial 2xy}{\partial x}~f_u(u,v)+(2xy)~\dfrac{\partial f_u(u,v)}{\partial x}\\&=(2y)f_u(u,v)+(2xy)(\dfrac{\partial u}{\partial x}f_{uu}(u,v)+\dfrac{\partial v}{\partial x}f_{uv}(u,v))\\&=2yf_u(xy^2,x^2)+2xy^3f_{uu}(xy^2,x^2)+4x^2yf_{uv}(xy^2,x^2)\end{align}$$

How do I compute the partial derivatives of f with respect to u and v? I have not been given any further information;

Therefore, while you have been promised that these derivatives exist, you cannot compute them.

So, that is all you can do and your job is done.

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