Problem 63 - Calculus Early Transcendentals

Question

Compute the first partial derivatives of the following functions. $$h(x, y, z)=(1+x+2 y)^{z}$$

Step-by-Step Solution

Answer

Question: Calculate the first partial derivatives of the function \(h(x, y, z) = (1+x+2y)^{z}\) with respect to x, y, and z. Answer: The first partial derivatives are: 1. \(\frac{\partial h}{\partial x}= z(1+x+2y)^{z-1}\) 2. \(\frac{\partial h}{\partial y}= 2z(1+x+2y)^{z-1}\) 3. \(\frac{\partial h}{\partial z}= h(x,y,z)\ln(1+x+2y)\)

Step 1: Calculate the partial derivative with respect to x, \(\frac{\partial h}{\partial x}\)

To find the partial derivative with respect to x, we treat y and z as constants. We have: $$\frac{\partial h}{\partial x}=(z(1+x+2y)^{z-1})(1)$$

Step 2: Calculate the partial derivative with respect to y, \(\frac{\partial h}{\partial y}\)

To find the partial derivative with respect to y, we treat x and z as constants. We have: $$\frac{\partial h}{\partial y}=(z(1+x+2y)^{z-1})(2)$$

Step 3: Calculate the partial derivative with respect to z, \(\frac{\partial h}{\partial z}\)

To find the partial derivative with respect to z, we treat x and y as constants. Therefore, we would need to apply the natural logarithm's chain rule since our variable z is in the exponent. We have: $$\frac{\partial h}{\partial z}= \frac{d(1+x+2y)^{z}}{dz}=h(x,y,z)\ln(1+x+2y)$$ So, the first partial derivatives of the function \(h(x, y, z) = (1+x+2y)^{z}\) are: 1. \(\frac{\partial h}{\partial x}=z(1+x+2y)^{z-1}\) 2. \(\frac{\partial h}{\partial y}=2z(1+x+2y)^{z-1}\) 3. \(\frac{\partial h}{\partial z}=h(x,y,z)\ln(1+x+2y)\)