Find \( z^{\prime}(t) \), where \( z=\frac{1}{x}+\frac{1}{y}, x=t^{2}+2 t \), and \( y=t^{3}-3 \), in the following ways. a. Replace \( x \) and \( y \) to write \( z \) as a function of \( t \) and differentiate. b. Use the Chain Rule. b. To find the derivative using the Chain Rule, begin by finding the intermediate derivatives. \( \frac{\partial z}{\partial x}=-\frac{1}{t^{2}}+2 t \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=2 t+2 \) (Type an expression using \( t \) as the variable.) \( \frac{\partial z}{\partial y}=-\frac{1}{y^{2}} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d y}{d t}=\square \) (Type an expression using \( t \) as the variable.)

To find \( z^{\prime}(t) \) where \( z = \frac{1}{x} + \frac{1}{y} \), \( x = t^2 + 2t \), and \( y = t^3 - 3 \), we will follow the steps outlined in the problem. ### a. Replace \( x \) and \( y \) to write \( z \) as a function of \( t \) and differentiate. First, we substitute \( x \) and \( y \) into the equation for \( z \): \[ z = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] Now, we differentiate \( z \) with respect to \( t \): \[ z^{\prime}(t) = \frac{d}{dt}\left(\frac{1}{t^2 + 2t}\right) + \frac{d}{dt}\left(\frac{1}{t^3 - 3}\right) \] Using the quotient rule, we find the derivatives: 1. For \( \frac{1}{t^2 + 2t} \): \[ \frac{d}{dt}\left(\frac{1}{t^2 + 2t}\right) = -\frac{(2t + 2)}{(t^2 + 2t)^2} \] 2. For \( \frac{1}{t^3 - 3} \): \[ \frac{d}{dt}\left(\frac{1}{t^3 - 3}\right) = -\frac{(3t^2)}{(t^3 - 3)^2} \] Combining these results, we have: \[ z^{\prime}(t) = -\frac{(2t + 2)}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} \] ### b. Use the Chain Rule. To find \( z^{\prime}(t) \) using the Chain Rule, we need to find the intermediate derivatives. 1. We already have: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] (Note: The expression provided in the prompt seems incorrect; it should be \( -\frac{1}{x^2} \) instead of \( -\frac{1}{t^2} + 2t \).) 2. We have: \[ \frac{d x}{d t} = 2t + 2 \] 3. For \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] 4. Now we need to find \( \frac{d y}{d t} \): \[ \frac{d y}{d t} = 3t^2 \] ### Summary of Intermediate Derivatives - \( \frac{\partial z}{\partial x} = -\frac{1}{x^2} \) - \( \frac{d x}{d t} = 2t + 2 \) - \( \frac{\partial z}{\partial y} = -\frac{1}{y^2} \) - \( \frac{d y}{d t} = 3t^2 \) ### Final Derivative Using Chain Rule Now we can use the Chain Rule to find \( z^{\prime}(t) \): \[ z^{\prime}(t) = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t} \] Substituting the intermediate derivatives: \[ z^{\prime}(t) = \left(-\frac{1}{x^2}\right)(2t + 2) + \left(-\frac{1}{y^2}\right)(3t^2) \] This gives us the expression for \( z^{\prime}(t) \) using the Chain Rule.