Problem 47 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(w, x, y, z)=w^{2} x y^{2}+x y^{3} z^{2}$$

Step-by-Step Solution

Answer

Question: Compute the first partial derivatives of the function $$f(w, x, y, z) = w^{2} x y^{2} + x y^{3} z^{2}$$ with respect to w, x, y, and z. Answer: The first partial derivatives of the function f are: $$\frac{\partial f}{\partial w} = 2wxy^2$$ $$\frac{\partial f}{\partial x} = w^2y^2 + y^3z^2$$ $$\frac{\partial f}{\partial y} = 2w^2xy + 3x y^{2} z^{2}$$ $$\frac{\partial f}{\partial z} = 2xyz^3y^3$$

Step 1: Compute the partial derivative with respect to w

To compute the partial derivative of f with respect to w, we will differentiate f with respect to w while treating x, y, and z as constants. $$\frac{\partial f}{\partial w} = \frac{\partial (w^{2} x y^{2} + x y^{3} z^{2})}{\partial w}$$ Differentiating with respect to w, we get: $$\frac{\partial f}{\partial w} = 2wxy^2$$

Step 2: Compute the partial derivative with respect to x

To compute the partial derivative of f with respect to x, we will differentiate f with respect to x while treating w, y, and z as constants. $$\frac{\partial f}{\partial x} = \frac{\partial (w^{2} x y^{2} + x y^{3} z^{2})}{\partial x}$$ Differentiating with respect to x, we get: $$\frac{\partial f}{\partial x} = w^2y^2 + y^3z^2$$

Step 3: Compute the partial derivative with respect to y

To compute the partial derivative of f with respect to y, we will differentiate f with respect to y while treating w, x, and z as constants. $$\frac{\partial f}{\partial y} = \frac{\partial (w^{2} x y^{2} + x y^{3} z^{2})}{\partial y}$$ Differentiating with respect to y, we get: $$\frac{\partial f}{\partial y} = 2w^2xy + 3x y^{2} z^{2}$$

Step 4: Compute the partial derivative with respect to z

To compute the partial derivative of f with respect to z, we will differentiate f with respect to z while treating w, x, and y as constants. $$\frac{\partial f}{\partial z} = \frac{\partial (w^{2} x y^{2} + x y^{3} z^{2})}{\partial z}$$ Differentiating with respect to z, we get: $$\frac{\partial f}{\partial z} = 2xyz^3y^3$$ In summary, we have the following first partial derivatives of the function f: $$\frac{\partial f}{\partial w} = 2wxy^2$$ $$\frac{\partial f}{\partial x} = w^2y^2 + y^3z^2$$ $$\frac{\partial f}{\partial y} = 2w^2xy + 3x y^{2} z^{2}$$ $$\frac{\partial f}{\partial z} = 2xyz^3y^3$$