Course:MATH200/HigherOrderChainRule

Chain rule for second order partial derivatives

Using the chain rule to compute second-order partial derivatives can be a little bit confusing. It helps to see an example to clarify it. This example is part of a problem in our text.

Example

Let z=f(x,y) be a function with continuous second-order partial derivatives. Suppose that

$x=e^s\cos t,y=e^s\sin t$

Compute $\frac{{\partial }^{2}z}{\partial s^2}$.

Solution

By the chain rule we have

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}=e^s\cos t\frac{\partial z}{\partial x}+e^s\sin t\frac{\partial z}{\partial y}.$

Now we have

 $\frac{{\partial }^{2}z}{\partial s^2}=\frac{\partial }{\partial s}\frac{\partial z}{\partial s}=e^s\cos t\frac{\partial z}{\partial x}+e^s\cos t\frac{\partial }{\partial s}\frac{\partial z}{\partial x}+e^s\sin t\frac{\partial z}{\partial y}+e^s\sin t\frac{\partial }{\partial s}\frac{\partial z}{\partial y}.$

We have to expand out the second and fourth terms now. We start by expanding

$\frac{\partial }{\partial s}\frac{\partial z}{\partial x}=\frac{{\partial }^{2}z}{\partial x^2}\frac{\partial x}{\partial s}+\frac{{\partial }^{2}z}{\partial y\partial x}\frac{\partial y}{\partial s}=e^s\cos t\frac{{\partial }^{2}z}{\partial x^2}+e^s\sin t\frac{{\partial }^{2}z}{\partial y\partial x}.$

Similarly

$\frac{\partial }{\partial s}\frac{\partial z}{\partial y}=\frac{{\partial }^{2}z}{\partial x\partial y}\frac{\partial x}{\partial s}+\frac{{\partial }^{2}z}{\partial y^2}\frac{\partial y}{\partial s}=e^s\cos t\frac{{\partial }^{2}z}{\partial x\partial y}+e^s\sin t\frac{{\partial }^{2}z}{\partial y^2}.$

Now we substitute to get

$\frac{{\partial }^{2}z}{\partial s^2}=e^s\cos t\frac{\partial z}{\partial x}+e^{2s}{\cos }^{}t\frac{{\partial }^{2}z}{\partial x^2}+e^{2s}\cos t\sin t\frac{{\partial }^{2}z}{\partial y\partial x}+e^s\sin t\frac{\partial z}{\partial y}+e^{2s}\sin t\cos t\frac{{\partial }^{2}z}{\partial x\partial y}+e^{2s}{\sin }^{}t\frac{{\partial }^{2}z}{\partial y^2}.$

Finally, we can simplify this a little bit using Clairaut's formula to

$\frac{{\partial }^{2}z}{\partial s^2}=e^s\cos t\frac{\partial z}{\partial x}+e^s\sin t\frac{\partial z}{\partial y}+e^{2s}{\cos }^{}t\frac{{\partial }^{2}z}{\partial x^2}+e^{2s}{\sin }^{}t\frac{{\partial }^{2}z}{\partial y^2}+2e^{2s}\sin t\cos t\frac{{\partial }^{2}z}{\partial x\partial y}.$