Problem 50 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the first partial derivatives of the function $$F(w, x, y, z) = w \sqrt{x + 2y + 3z}$$ with respect to $$w, x, y, z$$. Answer: $$\frac{\partial F}{\partial w} = \sqrt{x+2 y+3 z}$$ $$\frac{\partial F}{\partial x} = w \cdot \frac{1}{2\sqrt{x+2y+3z}}$$ $$\frac{\partial F}{\partial y} = w \cdot \frac{2}{2\sqrt{x+2y+3z}}$$ $$\frac{\partial F}{\partial z} = w \cdot \frac{3}{2\sqrt{x+2y+3z}}$$

Step 1: Find the partial derivative with respect to w

To find the partial derivative of $$F(w, x, y, z)$$ with respect to $$w$$, we treat $$x, y, z$$ as constants and differentiate the expression with respect to $$w$$. So we have: $$\frac{\partial F}{\partial w} = \frac{\partial}{\partial w} (w \sqrt{x+2 y+3 z})$$ Since $$x, y, z$$ are constants here, the partial derivative is simply: $$\frac{\partial F}{\partial w} = \sqrt{x+2 y+3 z}$$

Step 2: Find the partial derivative with respect to x

Now, we find the partial derivative of $$F(w, x, y, z)$$ with respect to $$x$$, treating $$w, y, z$$ as constants: $$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x} (w \sqrt{x+2 y+3 z})$$ For this, we will use the chain rule, which states that: $$\frac{\partial F}{\partial x} = w \frac{\partial}{\partial x} (\sqrt{x+2 y+3 z})$$ Differentiating the inside expression: $$\frac{\partial}{\partial x} (x+2 y+3 z) = 1$$ And differentiating the square root function with respect to its argument: $$\frac{1}{2\sqrt{x+2y+3z}}$$ Thus, we have: $$\frac{\partial F}{\partial x} = w \cdot 1 \cdot \frac{1}{2\sqrt{x+2y+3z}}$$

Step 3: Find the partial derivative with respect to y

Now, we find the partial derivative of $$F(w, x, y, z)$$ with respect to $$y$$, treating $$w, x, z$$ as constants: $$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y} (w \sqrt{x+2 y+3 z})$$ Using the chain rule, we have: $$\frac{\partial F}{\partial y} = w \frac{\partial}{\partial y} (\sqrt{x+2 y+3 z})$$ Differentiating the inside expression: $$\frac{\partial}{\partial y} (x+2 y+3 z) = 2$$ And applying the derivative to the square root function, we get: $$\frac{\partial F}{\partial y} = w \cdot 2 \cdot \frac{1}{2\sqrt{x+2y+3z}}$$

Step 4: Find the partial derivative with respect to z

Lastly, we find the partial derivative of $$F(w, x, y, z)$$ with respect to $$z$$, treating $$w, x, y$$ as constants: $$\frac{\partial F}{\partial z} = \frac{\partial}{\partial z} (w \sqrt{x+2 y+3 z})$$ Using the chain rule, we have: $$\frac{\partial F}{\partial z} = w \frac{\partial}{\partial z} (\sqrt{x+2 y+3 z})$$ Differentiating the inside expression: $$\frac{\partial}{\partial z} (x+2 y+3 z) = 3$$ And applying the derivative to the square root function, we get: $$\frac{\partial F}{\partial z} = w \cdot 3 \cdot \frac{1}{2\sqrt{x+2y+3z}}$$ So, the first partial derivatives for the given function $$F(w, x, y, z)$$ are: $$\frac{\partial F}{\partial w} = \sqrt{x+2 y+3 z}$$ $$\frac{\partial F}{\partial x} = w \cdot \frac{1}{2\sqrt{x+2y+3z}}$$ $$\frac{\partial F}{\partial y} = w \cdot \frac{2}{2\sqrt{x+2y+3z}}$$ $$\frac{\partial F}{\partial z} = w \cdot \frac{3}{2\sqrt{x+2y+3z}}$$