Problem 29 - Calculus Early Transcendentals

Question

Find the four second partial derivatives of the following functions. $$f(x, y)=y^{3} \sin 4 x$$

Step-by-Step Solution

Answer

Question: Find the four second partial derivatives of the given function: $$f(x,y) = y^3\sin(4x)$$ Answer: 1. \(\frac{\partial^2 f}{\partial x^2} = -16y^3\sin(4x)\) 2. \(\frac{\partial^2 f}{\partial y^2} = 6y\sin(4x)\) 3. \(\frac{\partial^2 f}{\partial x\partial y} = 12y^2\cos(4x)\) 4. \(\frac{\partial^2 f}{\partial y\partial x} = 12y^2\cos(4x)\)

Step 1: Find the first partial derivatives

Compute the first partial derivatives of the given function with respect to x and y: $$\frac{\partial f}{\partial x} = y^3\cos(4x) \cdot 4 = 4y^3\cos(4x)$$ $$\frac{\partial f}{\partial y} = 3y^2\sin(4x)$$

Step 2: Find the second partial derivatives

Now, compute the second partial derivatives by taking the derivative of the results from step 1 with respect to each variable: 1. \(\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(4y^3\cos(4x)) = -16y^3\sin(4x)\) 2. \(\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(3y^2\sin(4x)) = 6y\sin(4x)\) 3. \(\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial}{\partial y}(4y^3\cos(4x)) = 12y^2\cos(4x)\) 4. \(\frac{\partial^2 f}{\partial y\partial x} = \frac{\partial}{\partial x}(3y^2\sin(4x)) = 12y^2\cos(4x)\) From this, we can clearly see that the mixed second partial derivatives are equal: $$\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} = 12y^2\cos(4x)$$ The four second partial derivatives of the given function are: 1. \(\frac{\partial^2 f}{\partial x^2} = -16y^3\sin(4x)\) 2. \(\frac{\partial^2 f}{\partial y^2} = 6y\sin(4x)\) 3. \(\frac{\partial^2 f}{\partial x\partial y} = 12y^2\cos(4x)\) 4. \(\frac{\partial^2 f}{\partial y\partial x} = 12y^2\cos(4x)\)