Given \(f(x, y)=\sin 4 x \cos 3 y\) find \(\frac{\partial^{2} f}{\partial x^{2}}, \frac{\partial^{2} f}{\partial y^{2}}\) and \(\frac{\partial^{2} f}{\partial x \partial y}\)

First we need to find the first-order partial derivatives of the given function with respect to \(x\) and \(y\). The function \(f(x, y) = \sin 4x \cos 3y\) can be treated as the product of two functions, one only depending on \(x\) and the other only depending on \(y\). So let's compute these first-order partial derivatives. Using the chain rule for differentiation, we have $$\frac{\partial f}{\partial x} = \frac{d(\sin 4x)}{dx}\cos 3y = 4\cos 4x\cos 3y$$ and $$\frac{\partial f}{\partial y} = \sin 4x \frac{d(\cos 3y)}{dy} = -3\sin 4x\sin 3y$$

Now that we have the first-order partial derivatives, we can proceed to compute the second-order partial derivatives. To find \(\frac{\partial^2 f}{\partial x^2}\), differentiate \(\frac{\partial f}{\partial x}\) with respect to \(x\): $$\frac{\partial^2 f}{\partial x^2} = \frac{d(4\cos 4x\cos 3y)}{dx} = -16\sin 4x\cos 3y$$ To find \(\frac{\partial^2 f}{\partial y^2}\), differentiate \(\frac{\partial f}{\partial y}\) with respect to \(y\): $$\frac{\partial^2 f}{\partial y^2} = \frac{d(-3\sin 4x\sin 3y)}{dy} = -9\sin 4x\cos 3y$$ To find \(\frac{\partial^2 f}{\partial x\partial y}\), differentiate \(\frac{\partial f}{\partial x}\) with respect to \(y\): $$\frac{\partial^2 f}{\partial x\partial y} = \frac{d(4\cos 4x\cos 3y)}{dy} = -12\cos 4x\sin 3y$$ So, we have obtained the second-order partial derivatives of the function as: $$\frac{\partial^2 f}{\partial x^2} = -16\sin 4x\cos 3y\ ,\ \frac{\partial^2 f}{\partial y^2} = -9\sin 4x\cos 3y\ ,\ \text{ and }\ \frac{\partial^2 f}{\partial x\partial y} = -12\cos 4x\sin 3y.$$