Answer to Question #135766 in Calculus for Promise Omiponle

Question

5. Find all the first and second partial derivatives of f(x,y)=x^(2) y + cos(xy) + tan(x/y).

Answer 1

$\frac{\partial f}{\partial x}=2xy+y(-\sin(xy))+\frac{1}y\frac{1}{\cos^2(\frac{x}y)} \\ \frac{\partial f}{\partial y}=x^2+x(-\sin(xy))+(-\frac{x}{y^2})\frac{1}{\cos^2(\frac{x}y)}\\ \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial^2 f}{\partial y \partial x}=2x+(-\sin(xy))+xy(-\cos(xy))+\\+(-\frac{1}{y^2})\frac{1}{\cos^2(\frac{x}y)}+(-\frac{x}{y^2})(\frac{1}y)2\cos(\frac{x}y)(-\sin(\frac{x}{y}))(-\frac{1}{\cos^4(\frac{x}y)})=\\=2x-\sin(xy)-xy\cos(xy)-\frac{1}{y^2}\frac{1}{\cos^2(\frac{x}{y})}-\frac{x}{y^3}\frac{2\sin(\frac{x}y)}{\cos^3(\frac{x}y)} \\ \frac{\partial^2 f}{\partial x^2}=2y+y^2(-\cos(xy))+\frac{1}{y^2}(-\sin(\frac{x}{y}))(\frac{-2}{\cos^3(\frac{x}{y})})=\\=2y-y^2\cos(xy)+\frac{1}{y^2}\frac{2\sin(\frac{x}{y})}{\cos^3(\frac{x}{y})}\\ \frac{\partial^2 f}{\partial y^2}=-x^2\cos(xy)+\frac{2x}{y^3}\frac{1}{\cos^2(\frac{x}y)}+\frac{x^2}{y^4}\frac{2\sin(\frac{x}y)}{\cos^3(\frac{x}y)}$