Problem 23 - Calculus Early Transcendentals

Question

Find the following derivatives. \(z_{s}\) and \(z_{t},\) where \(z=e^{x+y}, x=s t,\) and \(y=s+t\)

Step-by-Step Solution

Answer

Answer: The partial derivatives of \(z = e^{x+y}\) with respect to \(s\) and \(t\) are: \(z_{s} = e^{x+y}(t+1)\) \(z_{t} = e^{x+y}(s+1)\)

Step 1: Compute the partial derivatives of x and y

To compute the partial derivatives of x = st and y = s+t with respect to s and t, we use the following formulas: \(\frac{\partial x}{\partial s} = t\) \(\frac{\partial x}{\partial t} = s\) \(\frac{\partial y}{\partial s} = 1\) \(\frac{\partial y}{\partial t} = 1\)

Step 2: Compute the partial derivatives of z

To compute the partial derivatives of z = e^(x+y) with respect to x and y, we use the following formulas: \(\frac{\partial z}{\partial x} = e^{x+y}\cdot \frac{\partial (x+y)}{\partial x} = e^{x+y}\) \(\frac{\partial z}{\partial y} = e^{x+y}\cdot \frac{\partial (x+y)}{\partial y} = e^{x+y}\)

Step 3: Apply the chain rule to find the partial derivatives of z with respect to s and t

Now, we will apply the chain rule to find the partial derivatives of z with respect to s and t. We will use the derivatives we computed in Steps 1 and 2: \(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial s} = e^{x+y} \cdot t + e^{x+y} \cdot 1 = e^{x+y}(t+1)\) \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial t} = e^{x+y} \cdot s + e^{x+y} \cdot 1 = e^{x+y}(s+1)\) So, the partial derivatives of z = e^(x+y) with respect to s and t are: \(z_{s} = e^{x+y}(t+1)\) \(z_{t} = e^{x+y}(s+1)\)