Consider the function $f(x,y)=\ln(x^2+y^2+3).$ Compute the partial derivatives of the first and second order.

Question

I'have been trying to calculate the last partial derivative but did not get the correct answer. My teacher's answer is $-4xy/(x^2+y^2+3)^2$. Can you please explain how we got it? (with a solution) I've been trying using the general quotient rule and the output was a total mess. My exam is in 5 hours :)

Thanks!

Answer 1

The notation $\frac{\partial^2f}{\partial x\partial y}$ is the same as $$ \frac{\partial}{\partial x}\frac{\partial f}{\partial y}= \frac{\partial}{\partial x}\frac{2y}{x^2+y^2+3}\\ =2y\frac{\partial}{\partial x}\frac{1}{x^2+y^2+3}=2y\frac{-2x}{(x^2+y^2+3)^2}\\ =\frac{-4xy}{(x^2+y^2+3)^2}\\ $$ by the quotient rule and noting that $y$ is constant when we take the partial in $x$.