Calculating second order partial derivative of a composite function

Question

Q. Given that $u=\arctan\left(\frac{x^3+y^3}{x-y}\right)$, prove the following :

$$x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=(1-4\sin^2 u)\sin(2u)$$

(The relevant partial derivatives are assumed to be continuous)

Attempted incomplete solution:

$$\tan(u)=\frac{x^3+y^3}{x-y}=f~\textrm{(say)}$$

We note that $f$ is a homogeneous function in $x,y$ of degree $2$ and hence, by a general result of Euler's Theorem, we have,

$$x^2\frac{\partial^2f}{\partial x^2}+2xy\frac{\partial^2f}{\partial x\partial y}+y^2\frac{\partial^2f}{\partial y^2}=2(2-1)f=2\tan(u)$$

I'm having trouble expressing the second order partial derivatives of $f$ in terms of that of $u$ (I'm relatively new at this). Can someone help me out? I don't want the complete solution, just how to apply the chain rule to get the partial derivatives of $f$ in terms of $u$. Thanks.

Answer 1

It's not a bad idea to keep track of your variable dependencies, especially when first learning chain rule problems. Specifically, your statement $\tan u = f$ is clearer when written $$\tan u(x,y) = f(x,y).$$ Now you can take the derivative of both sides with respect to $x$. $$\sec^2 (u(x,y)) \frac{\partial u}{\partial x}(x,y) = \frac{\partial f}{\partial x}(x,y).$$ Differentiating again gives $$2\sec^2 (u(x,y)) \tan (u(x,y)) \left(\frac{\partial u}{\partial x} (x,y) \right)^2 + \sec^2(u(x,y)) \frac{\partial^2 u}{\partial x^2}(x,y) = \frac{\partial ^2 f}{\partial x^2}(x,y).$$

It gets a little messy with all the dependencies, especially on the derivatives, so you can drop them when you feel comfortable doing so. But writing them out can help with chain rule issues.

Answer 2

Here is an outline of one way forward.

Designate the argument of the arctangent by the new variable $t$ so that

$$t(x,y)=\frac{x^3+y^3}{x-y}$$

Therefore, we can write $u(t(x,y))=\arctan (t(x,y))$. Then, we have

$$\begin{align} \sin u&=\frac{t}{\sqrt{1+t^2}} \tag 1\\\\ \sin(2u)&=\frac{2t}{1+t^2}\tag 2\\\\ u'(t)&=\frac{1}{1+t^2}\tag 3\\\\ u''(t)&=\frac{-2t}{(1+t^2)^2}\tag 4 \end{align}$$

And from the Chain Rule, we have

$$\begin{align} \frac{\partial u}{\partial x}&=u'(t)\frac{\partial t}{\partial x}\\\\ \frac{\partial u}{\partial y}&=u'(t)\frac{\partial t}{\partial y}\\\\ \frac{\partial^2 u}{\partial x^2}&=u''(t)\left(\frac{\partial t}{\partial x}\right)^2+u'(t)\frac{\partial^2 t}{\partial x^2} \tag 5\\\\ \frac{\partial^2 u}{\partial y^2}&=u''(t)\left(\frac{\partial t}{\partial y}\right)^2+u'(t)\frac{\partial^2 t}{\partial y^2} \tag 6\\\\ \frac{\partial^2 u}{\partial x \partial y}&=u''(t)\left(\frac{\partial t}{\partial x}\frac{\partial t}{\partial y}\right)+u'(t)\frac{\partial^2 t}{\partial x \partial y} \tag 7\\\\ \end{align}$$

Using $(5)-(7)$ and the general result of Euler's Theorem, we have

$$u''(t)\left(x\frac{\partial t}{\partial x}+y\frac{\partial t}{\partial y}\right)^2+2tu'(t)=(1-4\sin^2(u))\sin(2u)$$

Now, finish by calculating the partial derivatives of $t$ with respect to $x$ and $y$ and using $(1)-(4)$.