Verify $\large\frac{\partial ^{2} y}{\partial x\partial y}=\frac{\partial ^{2} y}{\partial y\partial x}$ for the following function;$\;u=\sin 3x\cos4y$

Question

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Answer 1

Toolbox:

• If $u=f(x,y)$ is the function of two independent variables then
• $\large\frac{\partial u}{\partial y (x_0,y_0)}=\frac{d}{dy}$$f(x_0,y)\;and \;\large\frac{\partial u}{\partial x(x_0,y_0)}=\large\frac{d}{dy}$$ f(x,y_0)$ Provided they exist
• The Second order parral derivaties are $\large\frac{\partial}{\partial x} \frac{\partial u}{\partial y}=\frac{\partial ^2 u}{\partial x \partial y},\frac{\partial}{\partial y} \frac{\partial u}{\partial y}=\frac{\partial ^2 u}{\partial y^2},\frac{\partial}{\partial y} \frac{\partial u}{\partial x}=\frac{\partial ^2 u}{\partial y \partial x},$$\;and\;\large\frac{\partial}{\partial x} \frac{\partial u}{\partial x}=\frac{\partial ^2 u}{\partial x ^2}$ Partial derivatives of functions of more similarly defined

Given $u=\sin 3x \cos 4y$

Step 1:

$\large\frac{\partial u}{\partial y}$$=-4 \sin 3x \sin 4y$

$\large\frac{\partial^2 u}{\partial x \partial y}$$=-12 \cos 3x \sin 4 y$

Step 2:

$\large\frac{\partial u}{\partial x}$$=3 \cos 3x \cos 4y$

$\large\frac{\partial^2 u}{\partial y \partial x}$$=-12 \cos 3x \sin 4 y$

Step 3:

$\large\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}$